CBSE Class 12 Mathematics HOTs Matrices and Determinants

1 Mark Questions

Q1 Write the number of possible matrices which can be made if it has 12 elements.

Matrices and Determinants

Points to Remember

• Matrix: A matrix is an ordered rectangular array of numbers or functions. These numbers or functions are called the elements or entries of the matrix.
• Order of a Matrix: A matrix with \( m \) rows and \( n \) columns is called a matrix of order \( m \times n \).
• Square Matrix: An \( m \times n \) matrix is called a square matrix of order \( n \) if \( m = n \).
• Column Matrix: A matrix with only one column is called a column matrix, i.e., \( A = [a_{ij}]_{m \times 1} \).
• Row Matrix: A matrix with only one row is called a row matrix, i.e., \( B = [b_{ij}]_{1 \times n} \).
• Zero Matrix: A matrix in which all elements are zero is called a zero matrix or null matrix.
• Diagonal Matrix: A square matrix is called a diagonal matrix if all its non-diagonal elements are zero.
• Scalar Matrix: A diagonal matrix in which all diagonal elements are equal is called a scalar matrix.
• Identity Matrix: A scalar matrix in which each diagonal element is 1 is called an identity matrix or unit matrix. It is denoted by \( I \).
  \( I = [e_{ij}]_{n \times n} \) where \( e_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \ne j \end{cases} \)
• Transpose of a Matrix: If \( A = [a_{ij}]_{m \times n} \) is an \( m \times n \) matrix, then the matrix obtained by interchanging the rows and columns of \( A \) is called the transpose of \( A \), denoted by \( A' \) or \( A^T \).
  Properties of Transpose:
  (i) \( (A')' = A \)
  (ii) \( (A + B)' = A' + B' \)
  (iii) \( (kA)' = kA' \), where \( k \) is a scalar
  (iv) \( (AB)' = B'A' \)
• Symmetric Matrix: A square matrix \( A = [a_{ij}] \) is symmetric if \( a_{ij} = a_{ji} \) for all \( i, j \). This means \( A' = A \).
• Skew-Symmetric Matrix: A square matrix \( A = [a_{ij}] \) is skew-symmetric if \( a_{ij} = -a_{ji} \) for all \( i, j \). This means \( A' = -A \). The diagonal elements of a skew-symmetric matrix are always zero.
• Determinant: To every square matrix \( A = [a_{ij}] \) of order \( n \times n \), we can associate a unique number (real or complex) called the determinant of \( A \), denoted by \( \text{det } A \) or \( |A| \).
  Properties:
  (i) \( |AB| = |A||B| \)
  (ii) \( |kA|_{n \times n} = k^n |A|_{n \times n} \), where \( k \) is a scalar.
• Area of a Triangle: The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
  \[ \Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]
• Collinearity: Three points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) are collinear if and only if:
  \[ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \]
• Adjoint of a Square Matrix: The adjoint of a square matrix \( A = [a_{ij}]_{n \times n} \) is the transpose of the matrix of cofactors, denoted by \( \text{adj } A \).
  If \( A = [a_{ij}]_{n \times n} \), then \( \text{adj } A = [A_{ji}]_{n \times n} \), where \( A_{ij} \) is the cofactor of \( a_{ij} \).
  Properties of Adjoint:
  (i) \( A(\text{adj } A) = (\text{adj } A)A = |A|I \)
  (ii) If \( A \) is a square matrix of order \( n \), then \( |\text{adj } A| = |A|^{n-1} \)
  (iii) \( \text{adj}(AB) = (\text{adj } B)(\text{adj } A) \)
• Singular and Non-Singular Matrices: A square matrix \( A \) is singular if \( |A| = 0 \). It is non-singular if \( |A| \ne 0 \).
• Inverse of a Matrix: If \( A \) is a non-singular square matrix, its inverse exists and is given by:
  \[ A^{-1} = \frac{1}{|A|} \text{adj } A \]
  Properties of Inverse:
  (i) \( AA^{-1} = A^{-1}A = I \)
  (ii) \( (A^{-1})^{-1} = A \)
  (iii) \( (AB)^{-1} = B^{-1}A^{-1} \)
  (iv) \( (A^T)^{-1} = (A^{-1})^T \)
• Solution of System of Linear Equations:
  For a system of equations represented as \( AX = B \):
  (i) If \( |A| \ne 0 \), the system is consistent and has a unique solution given by \( X = A^{-1}B \).
  (ii) If \( |A| = 0 \) and \( (\text{adj } A)B \ne 0 \), the system is inconsistent and has no solution.
  (iii) If \( |A| = 0 \) and \( (\text{adj } A)B = 0 \), the system is either consistent with infinitely many solutions or inconsistent with no solution.

Very Short Answer Type Questions (1 Mark)

Question 1. If \( \begin{bmatrix} x+3 & 4 \\ y-4 & x+y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix} \), find \( x \) and \( y \).
Answer:
Comparing the corresponding entries of the two equal matrices:
\( x + 3 = 5 \implies x = 2 \).
\( y - 4 = 3 \implies y = 7 \).
We can also verify using the remaining entry: \( x + y = 2 + 7 = 9 \), which is consistent with the matrix.
Therefore, we get \( x = 2 \) and \( y = 7 \).
In simple words: When two matrices are equal, their corresponding elements must match. Matching the components gives simple linear equations to solve for the variables.

Exam Tip: Always double check your solutions with any extra equations present in the matrix to confirm there are no algebraic inconsistencies.

Question 2. If \( A = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \), find \( AB \).
Answer:
Multiplying the two matrices:
\( AB = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} = \begin{bmatrix} i(0) + 0(i) & i(i) + 0(0) \\ 0(0) - i(i) & 0(i) - i(0) \end{bmatrix} \)
\( = \begin{bmatrix} 0 & i^2 \\ -i^2 & 0 \end{bmatrix} \).
Since \( i^2 = -1 \), we substitute this value:
\( AB = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \).
In simple words: Do standard matrix multiplication row by column, then use the complex number property that \( i^2 \) is equal to \( -1 \) to simplify the final matrix elements.

Exam Tip: Be very careful with minus signs when multiplying imaginary units, keeping in mind that \( -i^2 \) simplifies to positive 1.

Question 3. Find the value of \( a_{23} + a_{32} \) in the matrix \( A = [a_{ij}]_{3 \times 3} \) where \( a_{ij} = \begin{cases} |2i - j| & \text{if } i > j \\ -i + 2j + 3 & \text{if } i \le j \end{cases} \).
Answer:
First, find \( a_{23} \):
Here \( i = 2 \) and \( j = 3 \). Since \( 2 \le 3 \), we use the second case:
\( a_{23} = -2 + 2(3) + 3 = -2 + 6 + 3 = 7 \).
Next, find \( a_{32} \):
Here \( i = 3 \) and \( j = 2 \). Since \( 3 > 2 \), we use the first case:
\( a_{32} = |2(3) - 2| = |6 - 2| = 4 \).
Now, sum these two elements:
\( a_{23} + a_{32} = 7 + 4 = 11 \).
In simple words: Find the correct row and column values for both elements, pick the formula from the conditional rule based on whether row index is larger or smaller, calculate them, and add the results.

Exam Tip: Identifying the index relation (like \( i > j \) or \( i \le j \)) correctly is the main step. Double-check indices before applying the formula.

Question 4. If \( B \) be a \( 4 \times 5 \) type matrix, then what is the number of elements in the third column.
Answer:
A matrix of size \( 4 \times 5 \) has exactly 4 rows and 5 columns.
Since there are 4 rows, any column (including the third column) will contain exactly 4 elements.
In simple words: The number of rows in a matrix determines how many elements are present in each of its vertical columns.

Exam Tip: Remember that for an \( m \times n \) matrix, each column contains \( m \) elements and each row contains \( n \) elements.

Question 5. If \( A = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} \), find \( 3A - 2B \).
Answer:
First, find the scalar multiples of the matrices:
\( 3A = \begin{bmatrix} 15 & 6 \\ 0 & 27 \end{bmatrix} \)
\( 2B = \begin{bmatrix} 6 & 12 \\ 0 & -2 \end{bmatrix} \).
Now perform subtraction element-by-element:
\( 3A - 2B = \begin{bmatrix} 15-6 & 6-12 \\ 0-0 & 27-(-2) \end{bmatrix} = \begin{bmatrix} 9 & -6 \\ 0 & 29 \end{bmatrix} \).
In simple words: Multiply every element of matrix A by 3 and every element of matrix B by 2, and then subtract the corresponding values.

Exam Tip: Be careful with subtracting negative numbers, like \( 27 - (-2) \), which becomes addition.

Question 6. If \( A = \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 2 & -6 \end{bmatrix} \), find \( (A + B)' \).
Answer:
First, calculate the sum of matrices \( A \) and \( B \):
\( A + B = \begin{bmatrix} 2+1 & -3+0 \\ -7+2 & 5-6 \end{bmatrix} = \begin{bmatrix} 3 & -3 \\ -5 & -1 \end{bmatrix} \).
Now, find the transpose of this sum by interchanging rows and columns:
\( (A + B)' = \begin{bmatrix} 3 & -5 \\ -3 & -1 \end{bmatrix} \).
In simple words: Add the corresponding values of the matrices first, then turn the rows of that sum into columns to find the transpose.

Exam Tip: You can also find \( A' + B' \) directly as transpose distributes over addition; both methods yield the exact same result.

Question 7. If \( A = \begin{bmatrix} 1 & 0 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 \\ 5 \\ 6 \end{bmatrix} \), find \( AB \).
Answer:
Here, matrix \( A \) is of order \( 1 \times 3 \) and matrix \( B \) is of order \( 3 \times 1 \). Their product \( AB \) is of order \( 1 \times 1 \):
\( AB = \begin{bmatrix} 1(2) + 0(5) + 4(6) \end{bmatrix} = \begin{bmatrix} 2 + 0 + 24 \end{bmatrix} = \begin{bmatrix} 26 \end{bmatrix} \).
In simple words: Multiply the elements of the row with the elements of the column and sum them up to get a single-element matrix.

Exam Tip: Do not write the final answer as a scalar value (like 26); represent it as a matrix \( [26] \) since the multiplication of these two matrices yields a matrix.

Question 8. If \( A = \begin{bmatrix} 4 & x+2 \\ 2x-3 & x+1 \end{bmatrix} \) is symmetric matrix, then find \( x \).
Answer:
Since \( A \) is a symmetric matrix, its transpose \( A' \) must be equal to \( A \), which implies \( a_{12} = a_{21} \).
\( x + 2 = 2x - 3 \)
\( 2x - x = 2 + 3 \)
\( x = 5 \).
In simple words: For a matrix to be symmetric, the elements across the main diagonal must be mirror images of each other. Setting them equal helps solve for the variable.

Exam Tip: Equating non-diagonal symmetric positions is a direct way to solve matrix symmetric problems in 1-mark questions.

Question 9. For what value of \( x \) the matrix \( \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & -4 \\ 3 & 4 & x+5 \end{bmatrix} \) is skew symmetric matrix.
Answer:
For any skew-symmetric matrix, all the main diagonal elements must be equal to zero.
The element at the bottom-right diagonal position is \( x + 5 \).
Setting it to 0:
\( x + 5 = 0 \implies x = -5 \).
In simple words: The elements along the main diagonal of any skew-symmetric matrix are always 0. Setting the diagonal expression to 0 gives the answer.

Exam Tip: Remembering that diagonal elements of a skew-symmetric matrix are zero is a quick shortcut that saves you from calculating the full transpose.

Question 10. If \( A = \begin{bmatrix} 2 & 3 \\ 1 & 0 \end{bmatrix} = P + Q \) where \( P \) is symmetric and \( Q \) is skew-symmetric matrix, then find the matrix \( Q \).
Answer:
The skew-symmetric part of a square matrix \( A \) is given by the formula \( Q = \frac{1}{2}(A - A') \).
First, find the transpose of \( A \):
\( A' = \begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix} \).
Next, compute \( A - A' \):
\( A - A' = \begin{bmatrix} 2 & 3 \\ 1 & 0 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \).
Now, divide this difference by 2 to get \( Q \):
\( Q = \frac{1}{2} \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \).
In simple words: To find the skew-symmetric component, subtract the transpose of the matrix from the original matrix, and then divide every element of the result by 2.

Exam Tip: Verify your answer by checking if \( Q' = -Q \) and that its diagonal elements are indeed zero.

Question 11. Find the value of \( \begin{vmatrix} a + ib & c + id \\ -c + id & a - ib \end{vmatrix} \).
Answer:
Evaluating the determinant of the \( 2 \times 2 \) matrix:
\( \text{det} = (a + ib)(a - ib) - (c + id)(-c + id) \)
\( = (a^2 - i^2b^2) - ((id)^2 - c^2) \)
Since \( i^2 = -1 \):
\( = (a^2 + b^2) - (-d^2 - c^2) \)
\( = a^2 + b^2 + c^2 + d^2 \).
In simple words: Cross-multiply diagonal terms and subtract, then use complex algebra rules with \( i^2 = -1 \) to simplify the expression.

Exam Tip: Watch out for the sign of the second product term; the product of complex conjugates always results in a sum of squares.

Question 12. If \( \begin{vmatrix} 2x + 5 & 3 \\ 5x + 2 & 9 \end{vmatrix} = 0 \), find \( x \).
Answer:
Expand the determinant:
\( 9(2x + 5) - 3(5x + 2) = 0 \)
\( 18x + 45 - 15x - 6 = 0 \)
\( 3x + 39 = 0 \)
\( 3x = -39 \implies x = -13 \).
In simple words: Cross-multiply and subtract to write the determinant as an algebraic expression, then solve the linear equation for \( x \).

Exam Tip: Factoring out common numerical values from rows or columns before expanding can sometimes make the arithmetic calculations even simpler.

Question 13. For what value of \( k \), the matrix \( \begin{bmatrix} k & 2 \\ 3 & 4 \end{bmatrix} \) has no inverse.
Answer:
A square matrix has no inverse if its determinant is equal to zero (making it a singular matrix).
\( \begin{vmatrix} k & 2 \\ 3 & 4 \end{vmatrix} = 4k - 6 = 0 \)
\( 4k = 6 \implies k = \frac{3}{2} \).
In simple words: Calculate the determinant of the matrix, set it to zero, and solve the equation to find the value of \( k \).

Exam Tip: Always associate the phrase "has no inverse" or "invertible fails" directly with setting the determinant of the matrix to 0.

Question 14. If \( A = \begin{bmatrix} \sin 30^\circ & \cos 30^\circ \\ -\sin 60^\circ & \cos 60^\circ \end{bmatrix} \), what is \( |A| \).
Answer:
Calculate the determinant of \( A \):
\( |A| = \sin 30^\circ \cos 60^\circ - \cos 30^\circ (-\sin 60^\circ) \)
\( = \sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ \).
Using the trigonometric identity \( \sin(X+Y) = \sin X \cos Y + \cos X \sin Y \):
\( |A| = \sin(30^\circ + 60^\circ) = \sin 90^\circ = 1 \).
In simple words: Cross-multiply the terms of the determinant, which gives the standard trigonometric expansion for \( \sin(X+Y) \), then simplify to find the final numerical value.

Exam Tip: Recognizing trig identities within matrix expansions is highly elegant and saves you from substituting long fraction values.

Question 15. Find the cofactor of \( a_{12} \) in \( \begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix} \).
Answer:
The element \( a_{12} \) is at row 1, column 2. Its cofactor \( C_{12} \) is given by \( C_{12} = (-1)^{1+2} M_{12} = -M_{12} \).
Find the minor \( M_{12} \) by removing the first row and second column:
\( M_{12} = \begin{vmatrix} 6 & 4 \\ 1 & -7 \end{vmatrix} = 6(-7) - 4(1) = -42 - 4 = -46 \).
Thus, the cofactor is:
\( C_{12} = -(-46) = 46 \).
In simple words: Cross out the row and column of the given element, calculate the determinant of the remaining numbers, and multiply by \( -1 \) because the position sum is odd.

Exam Tip: Always write out the formula \( C_{ij} = (-1)^{i+j} M_{ij} \) to show that you applied the correct positional sign change.

Question 16. Find the minor of \( a_{23} \) in \( \begin{bmatrix} 1 & 3 & -2 \\ 4 & -5 & 6 \\ 3 & 5 & 2 \end{bmatrix} \).
Answer:
The element \( a_{23} \) lies in row 2, column 3.
Its minor \( M_{23} \) is the determinant obtained by crossing out row 2 and column 3:
\( M_{23} = \begin{vmatrix} 1 & 3 \\ 3 & 5 \end{vmatrix} = 1(5) - 3(3) = 5 - 9 = -4 \).
In simple words: Delete the second row and the third column, then evaluate the determinant of the remaining \( 2 \times 2 \) grid.

Exam Tip: Remember that minors do not include the positional sign factor \( (-1)^{i+j} \); that is only done when calculating cofactors.

Question 17. Find the value of \( P \), such that the matrix \( \begin{bmatrix} -1 & 2 \\ 4 & P \end{bmatrix} \) is singular.
Answer:
A matrix is singular when its determinant is equal to zero.
\( \begin{vmatrix} -1 & 2 \\ 4 & P \end{vmatrix} = -P - 8 = 0 \implies P = -8 \).
In simple words: Set the determinant of the given matrix to 0, write out the simple algebraic step, and solve for \( P \).

Exam Tip: Write down the definition "Singular matrix means determinant = 0" first to gain partial marks in case of an arithmetic slip.

Question 18. Find the value of \( x \) such that the points \( (0, 2) \), \( (1, x) \), and \( (3, 1) \) are collinear.
Answer:
For three coordinates to be collinear, the determinant formed by putting them into the triangle area equation must be zero:
\( \begin{vmatrix} 0 & 2 & 1 \\ 1 & x & 1 \\ 3 & 1 & 1 \end{vmatrix} = 0 \).
Expanding along the first row:
\( 0(x - 1) - 2(1 - 3) + 1(1 - 3x) = 0 \)
\( -2(-2) + 1 - 3x = 0 \)
\( 4 + 1 - 3x = 0 \)
\( 5 - 3x = 0 \implies x = \frac{5}{3} \).
In simple words: Put the coordinate points into the standard determinant matrix format, equate the expansion to zero, and solve the linear equation for \( x \).

Exam Tip: Standard coordinate geometry methods can also verify this, but using determinants is the expected method in this board chapter.

Question 19. Area of a triangle with vertices \( (k, 0) \), \( (1, 1) \), and \( (0, 3) \) is 5 unit. Find the value(s) of \( k \).
Answer:
The area of a triangle is given by the formula:
\( \text{Area} = \frac{1}{2} \left| \begin{vmatrix} k & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 3 & 1 \end{vmatrix} \right| = 5 \).
Multiplying by 2:
\( \left| \begin{vmatrix} k & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 3 & 1 \end{vmatrix} \right| = 10 \).
Expand the determinant along the first row:
\( k(1 - 3) - 0 + 1(3 - 0) = \pm 10 \)
\( -2k + 3 = \pm 10 \).
Case 1: \( -2k + 3 = 10 \implies -2k = 7 \implies k = -\frac{7}{2} \).
Case 2: \( -2k + 3 = -10 \implies -2k = -13 \implies k = \frac{13}{2} \).
Thus, \( k = -\frac{7}{2} \) or \( k = \frac{13}{2} \).
In simple words: Write out the determinant for the area of a triangle, multiply by 2, expand it, and set the value to both positive and negative 10 to find two possible coordinates.

Exam Tip: Never forget to use \( \pm \) for the area determinant value; area is always positive, so the absolute value yields two possible solutions.

Question 20. If \( A \) is a square matrix of order 3 and \( |A| = -2 \), find the value of \( |-3A| \).
Answer:
Using the determinant property \( |kA| = k^n |A| \), where \( n \) is the order of the square matrix:
\( |-3A| = (-3)^3 |A| \) (since \( n = 3 \))
\( = -27 \times (-2) = 54 \).
In simple words: When pulling a constant out of a determinant, raise the constant to the power of the matrix order, and then multiply by the original determinant value.

Exam Tip: A very common error is to just write \( -3 |A| \). Remember to apply the power rule \( k^n \) to avoid losing marks.

Question 21. If \( A = 2B \) where \( A \) and \( B \) are square matrices of order \( 3 \times 3 \) and \( |B| = 5 \), what is \( |A| \)?
Answer:
Using the determinant property \( |kB| = k^n |B| \), where \( n = 3 \):
\( |A| = |2B| = 2^3 |B| \)
\( = 8 \times 5 = 40 \).
In simple words: Since the matrix is of size \( 3 \times 3 \), multiplying it by 2 changes its determinant by a factor of \( 2^3 = 8 \). Multiplying this factor by 5 gives the answer.

Exam Tip: Always look at the order of the matrix given in the question description to identify the correct power value \( n \).

Question 22. What is the number of all possible matrices of order \( 2 \times 3 \) with each entry 0, 1 or 2.
Answer:
A matrix of size \( 2 \times 3 \) contains \( 2 \times 3 = 6 \) entry positions.
For each of the 6 positions, we have 3 choices of entries (0, 1, or 2).
Therefore, the total number of unique matrices is:
\( 3^6 = 729 \).
In simple words: Find the total number of boxes in the matrix, then calculate the choices raised to the power of the number of boxes.

Exam Tip: This permutation concept is a regular 1-mark question. Remember the basic formula: \( (\text{choices})^{(\text{number of elements})} \).

Question 23. Find the area of the triangle with vertices \( (0, 0) \), \( (6, 0) \) and \( (4, 3) \).
Answer:
Using the coordinates in the determinant area formula:
\( \text{Area} = \frac{1}{2} \left| \begin{vmatrix} 0 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix} \right| \).
Expanding along the first row:
\( \text{Area} = \frac{1}{2} \left| 1(6(3) - 0(4)) \right| = \frac{1}{2} |18| = 9 \text{ sq. units} \).
In simple words: Put the coordinate points into the determinant grid, expand along the easiest row containing zeros, and halve the result to get the positive area.

Exam Tip: Expanding along a row containing maximum zeros (like the first row here) is the fastest way to solve determinant calculations.

Question 24. If \( \begin{vmatrix} 2x & 4 \\ -1 & x \end{vmatrix} = \begin{vmatrix} 6 & -3 \\ 2 & 1 \end{vmatrix} \), find \( x \).
Answer:
Evaluate both determinants separately:
Left Side: \( (2x)(x) - (4)(-1) = 2x^2 + 4 \).
Right Side: \( 6(1) - (-3)(2) = 6 + 6 = 12 \).
Set them equal to solve:
\( 2x^2 + 4 = 12 \)
\( 2x^2 = 8 \)
\( x^2 = 4 \implies x = \pm 2 \).
In simple words: Cross-multiply and subtract on both sides to get an algebraic equation, simplify, and solve for \( x \).

Exam Tip: Do not just compare corresponding elements like you do in matrix equality. For determinants, you must fully evaluate both sides first.

Question 25. If \( A = \begin{bmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{bmatrix} \), write the value of det \( A \).
Answer:
Let \( |A| = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \).
Apply the row operation \( R_1 \to R_1 + R_2 \):
\( |A| = \begin{vmatrix} x+y+z & x+y+z & x+y+z \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \).
Take the common factor \( (x+y+z) \) out from the first row:
\( |A| = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \).
Since row 1 and row 3 are completely identical, the value of the determinant is 0:
\( |A| = (x+y+z) \times 0 = 0 \).
In simple words: Add the second row to the first row to make all elements in the top row equal to \( x+y+z \). Pulling this out leaves a row of 1s, matching the bottom row, which automatically makes the determinant 0.

Exam Tip: Identifying identical rows or columns is a powerful shortcut that immediately simplifies determinant proofs.

Question 26. If \( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \) such that \( |A| = -15 \), find \( a_{11} C_{21} + a_{12} C_{22} \) where \( C_{ij} \) is the cofactors of \( a_{ij} \) in \( A \).
Answer:
By the properties of determinants, if the elements of any row are multiplied by the cofactors of another row, their sum is always zero.
Here, the elements \( a_{11} \) and \( a_{12} \) belong to Row 1, whereas the cofactors \( C_{21} \) and \( C_{22} \) belong to Row 2.
Therefore, we have:
\( a_{11} C_{21} + a_{12} C_{22} = 0 \).
In simple words: Multiplying elements of one row with cofactors of a different row and adding them up always results in 0.

Exam Tip: Remember this property: summing products of elements of a row with their *own* cofactors gives \( |A| \), but with *another* row's cofactors, it is always 0.

Question 27. If \( A \) is a non-singular matrix of order 3 and \( |A| = -3 \) find \( |\text{adj } A| \).
Answer:
Using the standard determinant property of the adjoint:
\( |\text{adj } A| = |A|^{n-1} \), where \( n \) is the order of the square matrix.
Since \( n = 3 \) and \( |A| = -3 \):
\( |\text{adj } A| = (-3)^{3-1} = (-3)^2 = 9 \).
In simple words: Raise the determinant of the matrix to the power of \( (n-1) \) to find the determinant of its adjoint.

Exam Tip: Ensure you do not drop the negative sign before squaring. Even with negative determinants, the squared result will be positive.

Question 28. If \( A = \begin{bmatrix} 5 & -3 \\ 6 & 8 \end{bmatrix} \) find \( (\text{adj } A) \).
Answer:
For any \( 2 \times 2 \) matrix of the form \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the adjoint is found by swapping the main diagonal elements and changing the signs of the off-diagonal elements:
\( \text{adj } A = \begin{bmatrix} 8 & 3 \\ -6 & 5 \end{bmatrix} \).
In simple words: Swap the top-left and bottom-right numbers, and change the signs of the top-right and bottom-left numbers.

Exam Tip: This shortcut for \( 2 \times 2 \) adjoints is highly reliable and helps avoid calculating individual cofactors manually.

Question 29. Given a square matrix \( A \) of order \( 3 \times 3 \) such that \( |A| = 12 \) find the value of \( |A \text{ adj } A| \).
Answer:
We know the matrix identity:
\( A(\text{adj } A) = |A|I \).
Taking the determinant on both sides for a matrix of order 3:
\( |A \text{ adj } A| = ||A|I_3| = |A|^3 |I_3| = |A|^3 \).
Since \( |A| = 12 \):
\( |A \text{ adj } A| = 12^3 = 1728 \).
In simple words: The determinant of \( A \) times its adjoint is equal to the determinant of \( A \) raised to the power of its order.

Exam Tip: Remember that taking a scalar determinant of \( |A|I_n \) yields \( |A|^n \), where \( n \) is the order of the identity matrix.

Question 30. If \( A \) is a square matrix of order 3 such that \( |\text{adj } A| = 8 \) find \( |A| \).
Answer:
Using the relation:
\( |\text{adj } A| = |A|^{n-1} \), where \( n = 3 \).
\( 8 = |A|^{3-1} = |A|^2 \)
\( |A| = \pm \sqrt{8} = \pm 2\sqrt{2} \).
In simple words: Set the given determinant value equal to the square of \( |A| \), and solve to find both positive and negative roots.

Exam Tip: Always include the \( \pm \) symbol when taking the square root; both roots are mathematically correct and expected by examiners.

Question 31. Let \( A \) be a non-singular square matrix of order \( 3 \times 3 \) find \( |\text{adj } A| \) if \( |A| = 10 \).
Answer:
Using the standard identity for a matrix of order 3:
\( |\text{adj } A| = |A|^{3-1} = |A|^2 \).
Given \( |A| = 10 \):
\( |\text{adj } A| = 10^2 = 100 \).
In simple words: Square the determinant of the original matrix to find the determinant of its adjoint matrix.

Exam Tip: Write down the general formula \( |\text{adj } A| = |A|^{n-1} \) before substituting the values to ensure full conceptual marks.

Question 32. If \( A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \) find \( |(A^{-1})^{-1}| \).
Answer:
By the properties of matrix inverses:
\( (A^{-1})^{-1} = A \).
Therefore, we just need to find the determinant of \( A \):
\( |(A^{-1})^{-1}| = |A| = 2(4) - (-1)(3) = 8 + 3 = 11 \).
In simple words: The double inverse of a matrix returns the matrix to its original state, so we simply calculate the standard determinant of the original matrix.

Exam Tip: Utilizing basic matrix property simplifications first saves you from doing long matrix inversion steps.

Question 33. If \( A = \begin{bmatrix} -1 & 2 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 \\ -4 \\ 0 \end{bmatrix} \) find \( |AB| \).
Answer:
First, find the product matrix \( AB \):
\( AB = \begin{bmatrix} -1(3) + 2(-4) + 3(0) \end{bmatrix} = \begin{bmatrix} -3 - 8 + 0 \end{bmatrix} = \begin{bmatrix} -11 \end{bmatrix} \).
The determinant of a \( 1 \times 1 \) matrix is simply the value of its single element:
\( |AB| = -11 \).
In simple words: Multiply the elements of the row and column to get a single-element matrix, whose determinant is just that single value.

Exam Tip: Be sure to keep the negative sign in the final determinant calculation; unlike area, determinant values can certainly be negative.

Short Answer Type Questions (4 Marks)

Question 34. Find x, y, z and w if \( \begin{bmatrix} x-y & 2x+z \\ 2x-y & 3x+w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \).
Answer:
By comparing corresponding elements of the equal matrices, we form the following system of linear equations:
\( x - y = -1 \) --- (1)
\( 2x - y = 0 \implies y = 2x \) --- (2)
\( 2x + z = 5 \) --- (3)
\( 3x + w = 13 \) --- (4)
Substitute equation (2) into equation (1):
\( x - 2x = -1 \)
\( \implies -x = -1 \)
\( \implies x = 1 \).
Now substitute \( x = 1 \) into equation (2):
\( y = 2(1) = 2 \).
Substitute \( x = 1 \) into equation (3):
\( 2(1) + z = 5 \)
\( \implies z = 5 - 2 = 3 \).
Substitute \( x = 1 \) into equation (4):
\( 3(1) + w = 13 \)
\( \implies w = 13 - 3 = 10 \).
Therefore, the solution is \( x = 1, y = 2, z = 3, w = 10 \).
In simple words: Set the corresponding matrix parts equal to build four simple equations. Solve for \( x \) and \( y \) first, then use their values to find \( z \) and \( w \).

Exam Tip: Clearly label your equations (1), (2), etc., and show your substitution steps sequentially. This ensures you get full presentation marks.

Question 35. Construct a 3 x 3 matrix \( A = [a_{ij}] \) whose elements are given by \( a_{ij} = \begin{cases} \frac{1+i+j}{2} & \text{if } i \ge j \\ \frac{|i-2j|}{2} & \text{if } i < j \end{cases} \).
Answer:
Let us calculate the element \( a_{ij} \) for each position of a \( 3 \times 3 \) matrix:
For Row 1 (\( i = 1 \)):
\( a_{11} \) (since \( 1 \ge 1 \)): \( a_{11} = \frac{1+1+1}{2} = \frac{3}{2} \)
\( a_{12} \) (since \( 1 < 2 \multiangle \)): \( a_{12} = \frac{|1-2(2)|}{2} = \frac{|1-4|}{2} = \frac{3}{2} \)
\( a_{13} \) (since \( 1 < 3 \)): \( a_{13} = \frac{|1-2(3)|}{2} = \frac{|1-6|}{2} = \frac{5}{2} \)

For Row 2 (\( i = 2 \)):
\( a_{21} \) (since \( 2 \ge 1 \)): \( a_{21} = \frac{1+2+1}{2} = 2 \)
\( a_{22} \) (since \( 2 \ge 2 \)): \( a_{22} = \frac{1+2+2}{2} = \frac{5}{2} \)
\( a_{23} \) (since \( 2 < 3 \)): \( a_{23} = \frac{|2-2(3)|}{2} = \frac{|2-6|}{2} = 2 \)

For Row 3 (\( i = 3 \)):
\( a_{31} \) (since \( 3 \ge 1 \)): \( a_{31} = \frac{1+3+1}{2} = \frac{5}{2} \)
\( a_{32} \) (since \( 3 \ge 2 \)): \( a_{32} = \frac{1+3+2}{2} = 3 \)
\( a_{33} \) (since \( 3 \ge 3 \)): \( a_{33} = \frac{1+3+3}{2} = \frac{7}{2} \)

Thus, the constructed matrix is:
\( A = \begin{bmatrix} 3/2 & 3/2 & 5/2 \\ 2 & 5/2 & 2 \\ 5/2 & 3 & 7/2 \end{bmatrix} \).
In simple words: Find every single element of the \( 3 \times 3 \) grid by looking at its row and column coordinates. Apply the correct formula depending on which coordinate is larger, and compile them into the final matrix.

Exam Tip: Write down each element's calculation explicitly so that if there is a minor arithmetic error, you still earn partial marks.

Question 36. Find A and B if \( 2A + 3B = \begin{bmatrix} 1 & -2 & 3 \\ 2 & 0 & -1 \end{bmatrix} \) and \( A - 2B = \begin{bmatrix} 3 & 0 & 1 \\ -1 & 6 & 2 \end{bmatrix} \).
Answer:
We are given two equations:
\( 2A + 3B = \begin{bmatrix} 1 & -2 & 3 \\ 2 & 0 & -1 \end{bmatrix} \) --- (1)
\( A - 2B = \begin{bmatrix} 3 & 0 & 1 \\ -1 & 6 & 2 \end{bmatrix} \) --- (2)
Multiply equation (2) by 2:
\( 2A - 4B = \begin{bmatrix} 6 & 0 & 2 \\ -2 & 12 & 4 \end{bmatrix} \) --- (3)
Subtract equation (3) from equation (1):
\( (2A + 3B) - (2A - 4B) = \begin{bmatrix} 1 & -2 & 3 \\ 2 & 0 & -1 \end{bmatrix} - \begin{bmatrix} 6 & 0 & 2 \\ -2 & 12 & 4 \end{bmatrix} \)
\( \implies 7B = \begin{bmatrix} -5 & -2 & 1 \\ 4 & -12 & -5 \end{bmatrix} \)
\( \implies B = \begin{bmatrix} -5/7 & -2/7 & 1/7 \\ 4/7 & -12/7 & -5/7 \end{bmatrix} \).
Now, express \( A \) using equation (2):
\( A = 2B + \begin{bmatrix} 3 & 0 & 1 \\ -1 & 6 & 2 \end{bmatrix} \)
\( A = \begin{bmatrix} -10/7 & -4/7 & 2/7 \\ 8/7 & -24/7 & -10/7 \end{bmatrix} + \begin{bmatrix} 3 & 0 & 1 \\ -1 & 6 & 2 \end{bmatrix} \)
\( A = \begin{bmatrix} -10/7 + 3 & -4/7 + 0 & 2/7 + 1 \\ 8/7 - 1 & -24/7 + 6 & -10/7 + 2 \end{bmatrix} = \begin{bmatrix} 11/7 & -4/7 & 9/7 \\ 1/7 & 18/7 & 4/7 \end{bmatrix} \).
In simple words: This is solved like simultaneous linear equations. Multiply the second matrix equation by 2, subtract it from the first to find matrix \( B \), and then use \( B \) to find matrix \( A \).

Exam Tip: Be extra careful with fractions. Keep them in fraction form instead of decimals, as standard textbook solutions expect rational numbers.

Question 37. If \( A = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \) and \( B = \begin{bmatrix} -2 & -1 & -4 \end{bmatrix} \), verify that \( (AB)' = B'A' \).
Answer:
First, find the product \( AB \):
\( AB = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} -2 & -1 & -4 \end{bmatrix} = \begin{bmatrix} (-1)(-2) & (-1)(-1) & (-1)(-4) \\ 2(-2) & 2(-1) & 2(-4) \\ 3(-2) & 3(-1) & 3(-4) \end{bmatrix} = \begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{bmatrix} \).
Now, find the transpose of this product, \( (AB)' \):
\( (AB)' = \begin{bmatrix} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12 \end{bmatrix} \).
Next, compute the individual transposes \( B' \) and \( A' \):
\( B' = \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix} \) and \( A' = \begin{bmatrix} -1 & 2 & 3 \end{bmatrix} \).
Now multiply \( B' \) and \( A' \):
\( B'A' = \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix} \begin{bmatrix} -1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} (-2)(-1) & (-2)(2) & (-2)(3) \\ (-1)(-1) & (-1)(2) & (-1)(3) \\ (-4)(-1) & (-4)(2) & (-4)(3) \end{bmatrix} = \begin{bmatrix} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12 \end{bmatrix} \).
Comparing the two results, we see that:
\( (AB)' = B'A' \).
Hence verified.
In simple words: Multiply column \( A \) and row \( B \) to get a \( 3 \times 3 \) matrix, then transpose it. Separately transpose \( B \) and \( A \), multiply them, and show that both final matrices are completely identical.

Exam Tip: This is a fundamental property of matrices. Clearly show LHS and RHS derivations separately to make the proof easy to grade.

Question 38. Express the matrix \( \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} = P + Q \) where P is a symmetric and Q is a skew-symmetric matrix.
Answer:
Let \( A = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \).
The symmetric part is \( P = \frac{1}{2}(A + A') \) and the skew-symmetric part is \( Q = \frac{1}{2}(A - A') \).
First, find \( A' \):
\( A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \).
Now find the sum:
\( A + A' = \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} \).
Divide this by 2 to get \( P \):
\( P = \begin{bmatrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix} \).
Now find the difference:
\( A - A' = \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} \).
Divide this by 2 to get \( Q \):
\( Q = \begin{bmatrix} 0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{bmatrix} \).
Verify that \( P + Q = A \):
\( P + Q = \begin{bmatrix} 3+0 & 1/2+5/2 & -5/2+3/2 \\ 1/2-5/2 & -2+0 & -2+3 \\ -5/2-3/2 & -2-3 & 2+0 \end{bmatrix} = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} = A \).
Since \( P' = P \) and \( Q' = -Q \), the condition is satisfied.
In simple words: Find the transpose of the matrix. Calculate half of the sum of the matrix and its transpose to find \( P \), and half of the difference to find \( Q \). Show that adding \( P \) and \( Q \) returns the original matrix.

Exam Tip: Always show the check step \( P + Q = A \) at the end, as this is a key step examiners look for when awarding full marks.

Question 39. If \( A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \), then prove that \( A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix} \) where n is a natural number.
Answer:
We use the principle of mathematical induction to prove this statement.
For \( n = 1 \):
\( A^1 = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \), which is true.
Assume the statement holds true for \( n = k \):
\( A^k = \begin{bmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{bmatrix} \).
Now we prove it for \( n = k + 1 \):
\( A^{k+1} = A^k \cdot A = \begin{bmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{bmatrix} \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \)
\( = \begin{bmatrix} \cos k\theta \cos\theta - \sin k\theta \sin\theta & \cos k\theta \sin\theta + \sin k\theta \cos\theta \\ -\sin k\theta \cos\theta - \cos k\theta \sin\theta & -\sin k\theta \sin\theta + \cos k\theta \cos\theta \end{bmatrix} \).
Using the standard trigonometric formulas:
\( \cos(X+Y) = \cos X \cos Y - \sin X \sin Y \)
\( \sin(X+Y) = \sin X \cos Y + \cos X \sin Y \).
The matrix simplifies to:
\( A^{k+1} = \begin{bmatrix} \cos(k\theta + \theta) & \sin(k\theta + \theta) \\ -\sin(k\theta + \theta) & \cos(k\theta + \theta) \end{bmatrix} = \begin{bmatrix} \cos((k+1)\theta) & \sin((k+1)\theta) \\ -\sin((k+1)\theta) & \cos((k+1)\theta) \end{bmatrix} \).
Thus, the statement is true for \( n = k + 1 \). By mathematical induction, the result holds for all natural numbers \( n \).
In simple words: Prove that the rule works for \( n=1 \). Then, assuming it works for \( n=k \), multiply by matrix \( A \) one more time, use trigonometry addition formulas, and show that it works for \( n=k+1 \).

Exam Tip: Clearly write out the trigonometric formulas used during the step multiplication. This is a very common induction proof in exams.

Question 40. Let \( A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} \), \( C = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \), find a matrix D such that CD - AB = O.
Answer:
We are given \( CD - AB = O \implies CD = AB \).
First, calculate the product \( AB \):
\( AB = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} = \begin{bmatrix} 2(5) + (-1)(7) & 2(2) + (-1)(4) \\ 3(5) + 4(7) & 3(2) + 4(4) \end{bmatrix} = \begin{bmatrix} 10-7 & 4-4 \\ 15+28 & 6+16 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} \).
Since \( CD = AB \), we can multiply both sides by \( C^{-1} \) on the left:
\( D = C^{-1}(AB) \).
To find \( C^{-1} \), first find its determinant:
\( |C| = 2(8) - 5(3) = 16 - 15 = 1 \ne 0 \), so \( C \) is invertible.
Find the adjoint of \( C \):
\( \text{adj } C = \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix} \).
Thus, \( C^{-1} = \frac{1}{|C|} \text{adj } C = \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix} \).
Now calculate \( D \):
\( D = \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} = \begin{bmatrix} 8(3) + (-5)(43) & 8(0) + (-5)(22) \\ -3(3) + 2(43) & -3(0) + 2(22) \end{bmatrix} \)
\( = \begin{bmatrix} 24 - 215 & -110 \\ -9 + 86 & 44 \end{bmatrix} = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix} \).
In simple words: First calculate product \( AB \). Since we have \( CD = AB \), find the inverse of \( C \) and multiply it with the product matrix \( AB \) from the left to isolate and find \( D \).

Exam Tip: Be sure to multiply by \( C^{-1} \) on the left, i.e., \( D = C^{-1}(AB) \). Matrix multiplication is not commutative, so multiplying on the wrong side will lead to an incorrect answer.

Question 41. Find the value of x such that \( \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = 0 \).
Answer:
First, multiply the row matrix and the \( 3 \times 3 \) matrix:
\( \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 1(1) + x(2) + 1(15) & 1(3) + x(5) + 1(3) & 1(2) + x(1) + 1(2) \end{bmatrix} \)
\( = \begin{bmatrix} 16 + 2x & 6 + 5x & 4 + x \end{bmatrix} \).
Now multiply this row matrix by the final column matrix:
\( \begin{bmatrix} 16 + 2x & 6 + 5x & 4 + x \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = 0 \)
\( (16 + 2x)(1) + (6 + 5x)(2) + (4 + x)(x) = 0 \)
\( 16 + 2x + 12 + 10x + 4x + x^2 = 0 \)
\( x^2 + 16x + 28 = 0 \).
Solve the quadratic equation by factoring:
\( x^2 + 14x + 2x + 28 = 0 \)
\( x(x + 14) + 2(x + 14) = 0 \)
\( (x + 2)(x + 14) = 0 \).
Thus, the possible values are \( x = -2 \) or \( x = -14 \).
In simple words: Multiply the matrices from left to right to simplify the expression into a quadratic equation, then factor and solve to find the two possible values of \( x \).

Exam Tip: Be sure to keep intermediate terms in brackets to avoid basic sign errors during polynomial expansion.

Question 42. Prove that the product of the matrices \( \begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \\ \cos\theta\sin\theta & \sin^2\theta \end{bmatrix} \) and \( \begin{bmatrix} \cos^2\phi & \cos\phi\sin\phi \\ \cos\phi\sin\phi & \sin^2\phi \end{bmatrix} \) is the null matrix, when \( \theta \) and \( \phi \) differ by an odd multiple of \( \frac{\pi}{2} \).
Answer:
Let \( A = \begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \\ \cos\theta\sin\theta & \sin^2\theta \end{bmatrix} \) and \( B = \begin{bmatrix} \cos^2\phi & \cos\phi\sin\phi \\ \cos\phi\sin\phi & \sin^2\phi \end{bmatrix} \).
We find their product \( AB \):
\( AB = \begin{bmatrix} \cos^2\theta\cos^2\phi + \cos\theta\sin\theta\cos\phi\sin\phi & \cos^2\theta\cos\phi\sin\phi + \cos\theta\sin\theta\sin^2\phi \\ \cos\theta\sin\theta\cos^2\phi + \sin^2\theta\cos\phi\sin\phi & \cos\theta\sin\theta\cos\phi\sin\phi + \sin^2\theta\sin^2\phi \end{bmatrix} \).
We can factor out terms from each element:
- First entry: \( \cos\theta\cos\phi(\cos\theta\cos\phi + \sin\theta\sin\phi) = \cos\theta\cos\phi\cos(\theta - \phi) \)
- Second entry: \( \cos\theta\sin\phi(\cos\theta\cos\phi + \sin\theta\sin\phi) = \cos\theta\sin\phi\cos(\theta - \phi) \)
- Third entry: \( \sin\theta\cos\phi(\cos\theta\cos\phi + \sin\theta\sin\phi) = \sin\theta\cos\phi\cos(\theta - \phi) \)
- Fourth entry: \( \sin\theta\sin\phi(\cos\theta\cos\phi + \sin\theta\sin\phi) = \sin\theta\sin\phi\cos(\theta - \phi) \).
So, we can rewrite the product matrix as:
\( AB = \cos(\theta - \phi) \begin{bmatrix} \cos\theta\cos\phi & \cos\theta\sin\phi \\ \sin\theta\cos\phi & \sin\theta\sin\phi \end{bmatrix} \).
Since \( \theta \) and \( \phi \) differ by an odd multiple of \( \frac{\pi}{2} \), we can write:
\( \theta - \phi = (2k + 1)\frac{\pi}{2} \) for some \( k \in \mathbb{Z} \).
Thus, we have:
\( \cos(\theta - \phi) = \cos\left((2k+1)\frac{\pi}{2}\right) = 0 \).
Substituting this back:
\( AB = 0 \cdot \begin{bmatrix} \cos\theta\cos\phi & \cos\theta\sin\phi \\ \sin\theta\cos\phi & \sin\theta\sin\phi \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \), which is the null matrix.
Hence proved.
In simple words: Multiply the matrices, simplify using the formula \( \cos(\theta - \phi) \), and note that since the angle difference is an odd multiple of \( \pi/2 \), this cosine term is 0, which makes all elements of the matrix 0.

Exam Tip: Clearly show the factoring steps for each element before using the subtraction identity. This is where students often make algebraic slips.

Question 43. If \( A = \begin{bmatrix} 5 & 3 \\ 12 & 7 \end{bmatrix} \), show that \( A^2 - 12A - I = 0 \). Hence find \( A^{-1} \).
Answer:
First, find \( A^2 \):
\( A^2 = \begin{bmatrix} 5 & 3 \\ 12 & 7 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ 12 & 7 \end{bmatrix} = \begin{bmatrix} 5(5) + 3(12) & 5(3) + 3(7) \\ 12(5) + 7(12) & 12(3) + 7(7) \end{bmatrix} = \begin{bmatrix} 25+36 & 15+21 \\ 60+84 & 36+49 \end{bmatrix} = \begin{bmatrix} 61 & 36 \\ 144 & 85 \end{bmatrix} \).
Now compute \( A^2 - 12A - I \):
\( 12A = \begin{bmatrix} 60 & 36 \\ 144 & 84 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
\( A^2 - 12A - I = \begin{bmatrix} 61 & 36 \\ 144 & 85 \end{bmatrix} - \begin{bmatrix} 60 & 36 \\ 144 & 84 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 61-60-1 & 36-36-0 \\ 144-144-0 & 85-84-1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).
To find \( A^{-1} \), multiply the equation by \( A^{-1} \):
\( A^{-1}(A^2 - 12A - I) = O \)
\( A - 12I - A^{-1} = O \)
\( \implies A^{-1} = A - 12I \).
Substitute \( A \) and \( I \):
\( A^{-1} = \begin{bmatrix} 5 & 3 \\ 12 & 7 \end{bmatrix} - \begin{bmatrix} 12 & 0 \\ 0 & 12 \end{bmatrix} = \begin{bmatrix} -7 & 3 \\ 12 & -5 \end{bmatrix} \).
In simple words: Square the matrix, subtract 12 times the matrix and the identity matrix, and show it equals zero. Rearrange this verified equation to get \( A^{-1} = A - 12I \) and solve.

Exam Tip: Never calculate the inverse using the adjoint method when the question says "Hence find \( A^{-1} \)". You must use the matrix equation to earn full marks.

Short Answer Type Questions (4 Marks)

Question 44. If \( A = \begin{bmatrix} 2 & 3 \\ 4 & 7 \end{bmatrix} \), find f(A) where \( f(x) = x^2 - 5x - 2 \).
Answer:
To find \( f(A) \), we express the polynomial with matrix variables as follows:
\( f(A) = A^2 - 5A - 2I \).
First, let us calculate \( A^2 \):
\( A^2 = \begin{bmatrix} 2 & 3 \\ 4 & 7 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 4 & 7 \end{bmatrix} = \begin{bmatrix} 2(2)+3(4) & 2(3)+3(7) \\ 4(2)+7(4) & 4(3)+7(7) \end{bmatrix} = \begin{bmatrix} 4+12 & 6+21 \\ 8+28 & 12+49 \end{bmatrix} = \begin{bmatrix} 16 & 27 \\ 36 & 61 \end{bmatrix} \).
Next, find the matrix scalar multiple \( 5A \):
\( 5A = \begin{bmatrix} 10 & 15 \\ 20 & 35 \end{bmatrix} \).
Next, define the identity term \( 2I \) for a \( 2 \times 2 \) matrix:
\( 2I = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \).
Now, substitute these calculations into the function:
\( f(A) = \begin{bmatrix} 16 & 27 \\ 36 & 61 \end{bmatrix} - \begin{bmatrix} 10 & 15 \\ 20 & 35 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 16-10-2 & 27-15-0 \\ 36-20-0 & 61-35-2 \end{bmatrix} = \begin{bmatrix} 4 & 12 \\ 16 & 24 \end{bmatrix} \).
In simple words: Replace the variable \( x \) with matrix \( A \) and the constant 2 with \( 2I \). Calculate the square of the matrix, subtract 5 times the matrix and 2 times the identity matrix, and combine the terms.

Exam Tip: Never forget to attach the identity matrix \( I \) to the constant term when converting a scalar equation into a matrix equation; otherwise, the addition is undefined.

Question 45. If \( A = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \), find x and y such that \( A^2 - xA + yI = 0 \).
Answer:
First, let us calculate \( A^2 \):
\( A^2 = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 4(4)+3(2) & 4(3)+3(5) \\ 2(4)+5(2) & 2(3)+5(5) \end{bmatrix} = \begin{bmatrix} 16+6 & 12+15 \\ 8+10 & 6+25 \end{bmatrix} = \begin{bmatrix} 22 & 27 \\ 18 & 31 \end{bmatrix} \).
Now write the matrix equation:
\( \begin{bmatrix} 22 & 27 \\ 18 & 31 \end{bmatrix} - x \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} + y \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
Equating corresponding entries yields:
From the top-right entry:
\( 27 - 3x = 0 \implies 3x = 27 \implies x = 9 \).
From the bottom-left entry:
\( 18 - 2x = 0 \implies 2x = 18 \implies x = 9 \) (consistent).
From the top-left entry:
\( 22 - 4x + y = 0 \).
Substitute \( x = 9 \) into this equation:
\( 22 - 4(9) + y = 0 \)
\( 22 - 36 + y = 0 \)
\( \implies y = 14 \).
Let us verify with the bottom-right entry:
\( 31 - 5(9) + 14 = 31 - 45 + 14 = 0 \), which is satisfied.
Therefore, \( x = 9 \) and \( y = 14 \).
In simple words: Find \( A^2 \), substitute it into the polynomial equation along with \( A \) and \( I \), write out equations for individual elements, and solve to get \( x \) and \( y \).

Exam Tip: Always double check that the values of \( x \) and \( y \) you solve for satisfy all four individual element positions in the final zero matrix check.

Question 46. Find the matrix X so that \( X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \).
Answer:
Let \( A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \) of order \( 2 \times 3 \) and \( B = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \) of order \( 2 \times 3 \).
Since \( X \cdot A = B \), and \( A \) has 2 rows and 3 columns while \( B \) has 2 rows and 3 columns, the matrix \( X \) must be a square matrix of order \( 2 \times 2 \).
Let \( X = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \).
Substituting this into the equation:
\( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \).
By equating the corresponding elements:
For the first row:
\( a + 4b = -7 \) --- (1)
\( 2a + 5b = -8 \) --- (2)
Multiply (1) by 2:
\( 2a + 8b = -14 \) --- (3)
Subtract (2) from (3):
\( 3b = -6 \implies b = -2 \).
Substitute \( b = -2 \) in (1):
\( a + 4(-2) = -7 \implies a = 1 \).
We can verify with \( 3a + 6b = 3(1) + 6(-2) = -9 \), which is correct.

For the second row:
\( c + 4d = 2 \) --- (4)
\( 2c + 5d = 4 \) --- (5)
Multiply (4) by 2:
\( 2c + 8d = 4 \) --- (6)
Subtract (5) from (6):
\( 3d = 0 \implies d = 0 \).
Substitute \( d = 0 \) in (4):
\( c = 2 \).
We can verify with \( 3c + 6d = 3(2) + 6(0) = 6 \), which is correct.
Therefore, the matrix is \( X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix} \).
In simple words: Define \( X \) as a generic \( 2 \times 2 \) matrix with variable letters, multiply it with the first matrix, set up equations for each position, and solve them to find the numbers for \( X \).

Exam Tip: Show that the solved values satisfy the third column elements of the right-hand side matrix; this verifies your system of equations is correct.

Question 47. If \( A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} \), then show that \( (AB)^{-1} = B^{-1} A^{-1} \).
Answer:
Let us first find the product \( AB \):
\( AB = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 2(1)+3(-1) & 2(-2)+3(3) \\ 1(1)+(-4)(-1) & 1(-2)+(-4)(3) \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix} \).
Calculate the determinant of \( AB \):
\( |AB| = (-1)(-14) - 5(5) = 14 - 25 = -11 \).
Find the inverse matrix \( (AB)^{-1} \):
\( (AB)^{-1} = \frac{1}{-11} \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \).

Now, find the individual inverses \( A^{-1} \) and \( B^{-1} \):
For matrix \( A \):
\( |A| = 2(-4) - 3(1) = -11 \).
\( A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 4 & 3 \\ 1 & -2 \end{bmatrix} \).
For matrix \( B \):
\( |B| = 1(3) - (-2)(-1) = 1 \).
\( B^{-1} = \frac{1}{1} \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \).

Now, compute the product \( B^{-1} A^{-1} \):
\( B^{-1} A^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \left( \frac{1}{11} \begin{bmatrix} 4 & 3 \\ 1 & -2 \end{bmatrix} \right) = \frac{1}{11} \begin{bmatrix} 3(4)+2(1) & 3(3)+2(-2) \\ 1(4)+1(1) & 1(3)+1(-2) \end{bmatrix} \)
\( = \frac{1}{11} \begin{bmatrix} 12+2 & 9-4 \\ 4+1 & 3-2 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \).
Since \( (AB)^{-1} = B^{-1} A^{-1} \), the identity is verified.
In simple words: Find the product of \( A \) and \( B \) and calculate its inverse. Then find the inverses of \( A \) and \( B \) separately, multiply them in reverse order, and show that both calculations give the same matrix.

Exam Tip: Be careful with the order of multiplication when verifying this property. Remember that \( (AB)^{-1} \) is equal to \( B^{-1}A^{-1} \), not \( A^{-1}B^{-1} \).

Question 48. Test the consistency of the following system of equations by matrix method: \( 3x - y = 5 \); \( 6x - 2y = 3 \).
Answer:
We represent the system of equations in matrix form \( AX = B \):
\( A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \end{bmatrix} \), \( B = \begin{bmatrix} 5 \\ 3 \end{bmatrix} \).
Find the determinant of the coefficient matrix \( A \):
\( |A| = 3(-2) - (-1)(6) = -6 + 6 = 0 \).
Since \( |A| = 0 \), we must test the value of \( (\text{adj } A)B \).
Calculate the adjoint of \( A \):
\( \text{adj } A = \begin{bmatrix} -2 & 1 \\ -6 & 3 \end{bmatrix} \).
Now multiply \( (\text{adj } A) \) by matrix \( B \):
\( (\text{adj } A)B = \begin{bmatrix} -2 & 1 \\ -6 & 3 \end{bmatrix} \begin{bmatrix} 5 \\ 3 \end{bmatrix} = \begin{bmatrix} -2(5)+1(3) \\ -6(5)+3(3) \end{bmatrix} = \begin{bmatrix} -7 \\ -21 \end{bmatrix} \).
Since \( (\text{adj } A)B \ne O \), the system of equations is inconsistent and has no solution.
In simple words: Write the equations as matrices. The determinant of the coefficient matrix is 0, which means there is no unique solution. Since multiplying the adjoint by the constants does not equal 0, the lines are parallel and never meet.

Exam Tip: When \( |A| = 0 \), always explicitly state whether \( (\text{adj } A)B \) is equal to or not equal to the zero matrix to justify your consistency conclusion.

Question 49. Using elementary row transformations, find the inverse of the matrix \( A = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \), if possible.
Answer:
We set up the elementary row equation \( A = IA \):
\( \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A \).
Apply the row transformation \( R_1 \to R_1 + 3R_2 \):
The first row of the left-hand matrix becomes:
\( [6 + 3(-2) \quad -3 + 3(1)] = [0 \quad 0] \).
The equation becomes:
\( \begin{bmatrix} 0 & 0 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} A \).
Since all elements in the first row of the left-hand side matrix have become zero, the inverse of matrix \( A \) does not exist.
In simple words: Set up the row transformation equation. When you add three times the second row to the first row, the entire top row becomes 0, which proves the matrix is singular and cannot be inverted.

Exam Tip: If any row or column becomes entirely zeros during elementary transformations, stop and write a concluding statement that the inverse is not defined.

Question 50. By using elementary column transformation, find the inverse of \( A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} \).
Answer:
We set up the column transformation equation \( A = AI \):
\( \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} = A \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Swap column 1 and column 2 (\( C_1 \leftrightarrow C_2 \)):
\( \begin{bmatrix} 1 & 3 \\ 2 & 5 \end{bmatrix} = A \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \).
Perform the column operation \( C_2 \to C_2 - 3C_1 \):
\( \begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix} = A \begin{bmatrix} 0 & 1 \\ 1 & -3 \end{bmatrix} \).
Multiply column 2 by -1 (\( C_2 \to -C_2 \)):
\( \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} = A \begin{bmatrix} 0 & -1 \\ 1 & 3 \end{bmatrix} \).
Perform the column operation \( C_1 \to C_1 - 2C_2 \):
\( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = A \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix} \).
Therefore, the inverse is:
\( A^{-1} = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix} \).
In simple words: Use step-by-step column operations to turn the left-hand matrix into the identity matrix \( I \). Applying those same operations to the right-hand matrix transforms it into the inverse.

Exam Tip: When using column transformations, always write \( A = AI \) (and not \( A = IA \)), and make sure you only perform column-level operations.

Question 51. If \( A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \) and \( A + A' = I \), then find the general value of \( \alpha \).
Answer:
First, find the transpose \( A' \):
\( A' = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \).
Compute the sum \( A + A' \):
\( A + A' = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} \).
Set this sum equal to the identity matrix \( I \):
\( \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
By equating the corresponding elements:
\( 2\cos\alpha = 1 \implies \cos\alpha = \frac{1}{2} \).
Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \), the general trigonometric solution is:
\( \alpha = 2n\pi \pm \frac{\pi}{3} \), where \( n \in \mathbb{Z} \).
In simple words: Transpose the matrix, add it to the original, set the diagonal element equation equal to 1, and solve the cosine equation to find the general angle formula.

Exam Tip: Always state the general solution formula (which includes \( 2n\pi \)) rather than just the principal value angle to secure full marks.

Question 52. Prove the following: \( \begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a+b+c)^3 \).
Answer:
Let \( \Delta \) be the determinant.
Perform the row operation \( R_1 \to R_1 + R_2 + R_3 \):
The first row elements become:
\( R_{11} = a-b-c + 2b + 2c = a+b+c \)
\( R_{12} = 2a + b-c-a + 2c = a+b+c \)
\( R_{13} = 2a + 2b + c-a-b = a+b+c \).
Our determinant becomes:
\( \Delta = \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \).
Take the common factor \( (a+b+c) \) out from the first row:
\( \Delta = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \).
Now perform the column operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = (a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ 2b & -b-c-a & 0 \\ 2c & 0 & -c-a-b \end{vmatrix} \).
Taking negative signs out of columns 2 and 3:
\( \Delta = (a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ 2b & -(a+b+c) & 0 \\ 2c & 0 & -(a+b+c) \end{vmatrix} \).
Expanding along the first row:
\( \Delta = (a+b+c) \cdot 1 \cdot \left[ (-(a+b+c)) \cdot (-(a+b+c)) - 0 \right] \)
\( = (a+b+c) \cdot (a+b+c)^2 = (a+b+c)^3 \).
Hence proved.
In simple words: Add all rows to the top row, which creates a common term \( a+b+c \) that can be factored out. Subtract the first column from the others to create zeros, making it easy to expand and prove the cubic term.

Exam Tip: Applying transformations that create multiple zeros in a single row or column is the most effective way to expand complex determinants.

Question 53. Prove the following: \( \begin{vmatrix} x+2 & x+3 & x+2a \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{vmatrix} = 0 \) if a, b, c are in A.P.
Answer:
Since \( a, b, c \) are in Arithmetic Progression (A.P.), we have the relation:
\( b - a = c - b \implies 2b = a + c \).
Let our determinant be \( \Delta \).
Apply the row operation \( R_1 \to R_1 + R_3 \):
\( \Delta = \begin{vmatrix} (x+2)+(x+4) & (x+3)+(x+5) & (x+2a)+(x+2c) \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{vmatrix} \)
\( = \begin{vmatrix} 2x+6 & 2x+8 & 2x+2(a+c) \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{vmatrix} \).
Substitute \( a+c = 2b \) into the first row elements:
\( \Delta = \begin{vmatrix} 2x+6 & 2x+8 & 2x+4b \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{vmatrix} \).
Factor out the number 2 from the first row:
\( \Delta = 2 \begin{vmatrix} x+3 & x+4 & x+2b \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{vmatrix} \).
Since row 1 and row 2 are completely identical, the value of the determinant is 0:
\( \Delta = 2 \times 0 = 0 \).
Hence proved.
In simple words: Since \( a, b, c \) are in A.P., we know \( 2b = a + c \). Add the third row to the first, substitute the A.P. identity, and factor out 2 to make the first two rows identical, which makes the determinant 0.

Exam Tip: Always state the algebraic property of the A.P. progression \( 2b = a + c \) at the beginning of your proof steps to show your logical flow.

Question 54. Prove the following: \( \begin{vmatrix} \sin\alpha & \cos\alpha & \sin(\alpha + \delta) \\ \sin\beta & \cos\beta & \sin(\beta + \delta) \\ \sin\gamma & \cos\gamma & \sin(\gamma + \delta) \end{vmatrix} = 0 \).
Answer:
Using the angle sum identity \( \sin(X+Y) = \sin X \cos Y + \cos X \sin Y \), expand the third column:
\( \Delta = \begin{vmatrix} \sin\alpha & \cos\alpha & \sin\alpha\cos\delta + \cos\alpha\sin\delta \\ \sin\beta & \cos\beta & \sin\beta\cos\delta + \cos\beta\sin\delta \\ \sin\gamma & \cos\gamma & \sin\gamma\cos\delta + \cos\gamma\sin\delta \end{vmatrix} \).
Perform the column operation \( C_3 \to C_3 - \cos\delta \cdot C_1 \):
\( \Delta = \begin{vmatrix} \sin\alpha & \cos\alpha & \cos\alpha\sin\delta \\ \sin\beta & \cos\beta & \cos\beta\sin\delta \\ \sin\gamma & \cos\gamma & \cos\gamma\sin\delta \end{vmatrix} \).
Now perform the column operation \( C_3 \to C_3 - \sin\delta \cdot C_2 \):
\( \Delta = \begin{vmatrix} \sin\alpha & \cos\alpha & 0 \\ \sin\beta & \cos\beta & 0 \\ \sin\gamma & \cos\gamma & 0 \end{vmatrix} \).
Since the third column consists entirely of zeros, the value of the determinant is 0:
\( \Delta = 0 \).
Hence proved.
In simple words: Expand the third column using trig identities. Use column operations to subtract the trigonometric multiples of the first two columns from the third column, reducing it to a column of zeros.

Exam Tip: Expanding a single column using sum rules allows you to eliminate dependent components systematically without expanding the entire matrix.

Short Answer Type Questions (4 Marks)

Question 55. Prove that \( \begin{vmatrix} b^2+c^2 & a^2 & a^2 \\ b^2 & c^2+a^2 & b^2 \\ c^2 & c^2 & a^2+b^2 \end{vmatrix} = 4a^2 b^2 c^2 \).
Answer:
Let \( \Delta \) be the given determinant.
Apply the row operation \( R_1 \to R_1 - R_2 - R_3 \):
- The first element of row 1 becomes: \( (b^2+c^2) - b^2 - c^2 = 0 \).
- The second element of row 1 becomes: \( a^2 - (c^2+a^2) - c^2 = -2c^2 \).
- The third element of row 1 becomes: \( a^2 - b^2 - (a^2+b^2) = -2b^2 \).
So, we get:
\( \Delta = \begin{vmatrix} 0 & -2c^2 & -2b^2 \\ b^2 & c^2 + a^2 & b^2 \\ c^2 & c^2 & a^2 + b^2 \end{vmatrix} \).
Expanding this along the first row:
\( \Delta = 0 - (-2c^2) \begin{vmatrix} b^2 & b^2 \\ c^2 & a^2 + b^2 \end{vmatrix} + (-2b^2) \begin{vmatrix} b^2 & c^2 + a^2 \\ c^2 & c^2 \end{vmatrix} \)
\( = 2c^2 \left[ b^2(a^2 + b^2) - b^2 c^2 \right] - 2b^2 \left[ b^2 c^2 - c^2(c^2 + a^2) \right] \)
\( = 2c^2 \left[ a^2 b^2 + b^4 - b^2 c^2 \right] - 2b^2 \left[ b^2 c^2 - c^4 - a^2 c^2 \right] \)
\( = 2a^2 b^2 c^2 + 2b^4 c^2 - 2b^2 c^4 - 2b^4 c^2 + 2b^2 c^4 + 2a^2 b^2 c^2 \)
\( = 4a^2 b^2 c^2 \).
Hence proved.
In simple words: Apply a row operation to create a zero and factorable terms in the first row. Expand along that first row to easily simplify the algebraic terms and prove the identity.

Exam Tip: Row subtraction is a standard first step for proving sum-of-squares determinants. Look to eliminate the squared sums in diagonal entries first.

Question 56. Prove that \( \begin{vmatrix} b+c & c+a & a+b \\ q+r & r+p & p+q \\ y+z & z+x & x+y \end{vmatrix} = 2 \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} \).
Answer:
Let \( \Delta \) be the determinant on the left-hand side.
Apply the column operation \( C_1 \to C_1 + C_2 + C_3 \):
\( \Delta = \begin{vmatrix} 2(a+b+c) & c+a & a+b \\ 2(p+q+r) & r+p & p+q \\ 2(x+y+z) & z+x & x+y \end{vmatrix} \).
Factor out the number 2 from the first column:
\( \Delta = 2 \begin{vmatrix} a+b+c & c+a & a+b \\ p+q+r & r+p & p+q \\ x+y+z & z+x & x+y \end{vmatrix} \).
Now perform the column operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = 2 \begin{vmatrix} a+b+c & -b & -c \\ p+q+r & -q & -r \\ x+y+z & -y & -z \end{vmatrix} \).
Perform the column operation \( C_1 \to C_1 + C_2 + C_3 \):
\( \Delta = 2 \begin{vmatrix} a & -b & -c \\ p & -q & -r \\ x & -y & -z \end{vmatrix} \).
Finally, factor out \( -1 \) from both the second and third columns:
\( \Delta = 2 \times (-1) \times (-1) \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} = 2 \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} \).
Hence proved.
In simple words: Add all columns to the first one to create a factor of 2, pull it out, and then subtract the columns systematically to isolate individual variables in each position.

Exam Tip: Be mindful of sign changes; when pulling out negative factors from multiple columns, multiply them together to confirm the final sign.

Question 57. Prove that \( \begin{vmatrix} a^2 & bc & ac+c^2 \\ a^2+ab & b^2 & ac \\ ab & b^2+bc & c^2 \end{vmatrix} = 4a^2 b^2 c^2 \).
Answer:
Let \( \Delta \) be the determinant.
Factor out \( a \) from column 1, \( b \) from column 2, and \( c \) from column 3:
\( \Delta = abc \begin{vmatrix} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{vmatrix} \).
Perform the row operation \( R_1 \to R_1 - R_2 - R_3 \):
\( \Delta = abc \begin{vmatrix} -2b & -2b & 0 \\ a+b & b & a \\ b & b+c & c \end{vmatrix} \).
Factor out the common term \( -2b \) from the first row:
\( \Delta = -2ab^2c \begin{vmatrix} 1 & 1 & 0 \\ a+b & b & a \\ b & b+c & c \end{vmatrix} \).
Apply column operation \( C_2 \to C_2 - C_1 \):
\( \Delta = -2ab^2c \begin{vmatrix} 1 & 0 & 0 \\ a+b & -a & a \\ b & c & c \end{vmatrix} \).
Now expand along the first row:
\( \Delta = -2ab^2c \cdot 1 \cdot \begin{vmatrix} -a & a \\ c & c \end{vmatrix} \)
\( = -2ab^2c \left[ -ac - ac \right] = -2ab^2c (-2ac) = 4a^2 b^2 c^2 \).
Hence proved.
In simple words: Pull out row factors \( a,b,c \), use subtraction to create zeros in the first row, simplify with column operations, and expand to find the identity.

Exam Tip: Extracting common variables from columns first is a crucial step that reduces quadratic matrix entries to linear ones.

Question 58. Prove that \( \begin{vmatrix} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{vmatrix} = x^2(x+a+b+c) \).
Answer:
Let \( \Delta \) be the determinant.
Perform the column operation \( C_1 \to C_1 + C_2 + C_3 \):
\( \Delta = \begin{vmatrix} x+a+b+c & b & c \\ x+a+b+c & x+b & c \\ x+a+b+c & b & x+c \end{vmatrix} \).
Factor out the term \( (x+a+b+c) \) from the first column:
\( \Delta = (x+a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & x+b & c \\ 1 & b & x+c \end{vmatrix} \).
Now apply the row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = (x+a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & x & 0 \\ 0 & 0 & x \end{vmatrix} \).
Expanding along the first column:
\( \Delta = (x+a+b+c) \cdot 1 \cdot \left[ x^2 - 0 \right] = x^2(x+a+b+c) \).
Hence proved.
In simple words: Sum all columns into the first column to create the factor \( (x+a+b+c) \). Pull it out, create zeros in the lower rows, and expand the remaining upper triangular determinant.

Exam Tip: When the sum of all elements in each row is the same, adding columns is always the quickest way to find a common factor.

Question 59. Show that : \( \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ yz & zx & xy \end{vmatrix} = (y-z)(z-x)(x-y)(yz+zx+xy) \).
Answer:
Multiply column 1 by \( x \), column 2 by \( y \), and column 3 by \( z \), and divide the determinant by \( xyz \):
\( \Delta = \frac{1}{xyz} \begin{vmatrix} x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \\ xyz & xyz & xyz \end{vmatrix} \).
Factor out \( xyz \) from row 3:
\( \Delta = \frac{xyz}{xyz} \begin{vmatrix} x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \\ 1 & 1 & 1 \end{vmatrix} = \begin{vmatrix} x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \\ 1 & 1 & 1 \end{vmatrix} \).
Now perform column operations \( C_1 \to C_1 - C_2 \) and \( C_2 \to C_2 - C_3 \):
\( \Delta = \begin{vmatrix} x^2 - y^2 & y^2 - z^2 & z^2 \\ x^3 - y^3 & y^3 - z^3 & z^3 \\ 0 & 0 & 1 \end{vmatrix} \).
Factor out \( (x-y) \) from column 1 and \( (y-z) \) from column 2:
\( \Delta = (x-y)(y-z) \begin{vmatrix} x+y & y+z & z^2 \\ x^2+xy+y^2 & y^2+yz+z^2 & z^3 \\ 0 & 0 & 1 \end{vmatrix} \).
Expanding along the third row:
\( \Delta = (x-y)(y-z) \left[ (x+y)(y^2+yz+z^2) - (y+z)(x^2+xy+y^2) \right] \).
Simplifying the term inside the bracket:
\( (x+y)(y^2+yz+z^2) - (y+z)(x^2+xy+y^2) \)
\( = xy^2 + xyz + xz^2 + y^3 + y^2z + yz^2 - (yx^2 + xy^2 + y^3 + zx^2 + xyz + zy^2) \)
\( = xz^2 + yz^2 - yx^2 - zx^2 \)
\( = z^2(y-x) + z(x^2-y^2) - yx(x-y) \)
\( = (z-x)(yz+zx+xy) \).
Substituting this back gives:
\( \Delta = (y-z)(z-x)(x-y)(yz+zx+xy) \).
Hence proved.
In simple words: Scale the columns with variables to find a common row term, apply difference operations to factor out the cyclic differences, and expand the remaining terms.

Exam Tip: Multiplying rows/columns by variables and dividing the determinant is a standard technique for handling cyclic determinant proofs.

Question 60. (i) If the points (a, b) (a´, b´) and (a - a´, b - b´) are collinear. Show that ab´ = a´b.
(ii) If \( A = \begin{bmatrix} 2 & 5 \\ 2 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & -3 \\ 2 & 5 \end{bmatrix} \), verify that \( |AB| = |A||B| \).
Answer:
(i) Since the given points are collinear, the determinant of their coordinates is zero:
\( \begin{vmatrix} a & b & 1 \\ a' & b' & 1 \\ a - a' & b - b' & 1 \end{vmatrix} = 0 \).
Apply row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \begin{vmatrix} a & b & 1 \\ a' - a & b' - b & 0 \\ -a' & -b' & 0 \end{vmatrix} = 0 \).
Expanding along the third column:
\( 1 \left[ (a' - a)(-b') - (b' - b)(-a') \right] = 0 \)
\( -a'b' + ab' - (-a'b' + a'b) = 0 \)
\( ab' - a'b = 0 \implies ab' = a'b \).
Hence proved.

(ii) First, calculate the individual determinants:
\( |A| = 2(1) - 5(2) = 2 - 10 = -8 \).
\( |B| = 4(5) - (-3)(2) = 20 + 6 = 26 \).
\( |A||B| = (-8)(26) = -208 \).
Now find the product matrix \( AB \):
\( AB = \begin{bmatrix} 2 & 5 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 4 & -3 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 2(4)+5(2) & 2(-3)+5(5) \\ 2(4)+1(2) & 2(-3)+1(5) \end{bmatrix} = \begin{bmatrix} 18 & 19 \\ 10 & -1 \end{bmatrix} \).
Calculate the determinant of the product:
\( |AB| = 18(-1) - 19(10) = -18 - 190 = -208 \).
Since \( |AB| = |A||B| \), the property is verified.
In simple words: For part (i), setting the collinearity coordinate determinant to 0 directly yields the algebraic relation. For part (ii), compute the product matrix first, and verify its determinant equals the product of individual determinants.

Exam Tip: Be precise when expanding the collinearity determinant; minor sign mistakes are the most common source of errors in coordinate geometry proofs.

Question 61. Given \( A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -2 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{bmatrix} \). Find the product AB and also find \( (AB)^{-1} \).
Answer:
First, calculate the product \( AB \):
\( AB = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -2 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 0(0)+(-1)(1)+2(1) & 0(1)+(-1)(0)+2(1) \\ 2(0)+(-2)(1)+0(1) & 2(1)+(-2)(0)+0(1) \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 2 \\ -2 & 2 \end{bmatrix} \).
Now find the determinant of the product:
\( |AB| = 1(2) - 2(-2) = 2 + 4 = 6 \).
Since \( |AB| \ne 0 \), the inverse matrix exists and is given by:
\( (AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB) = \frac{1}{6} \begin{bmatrix} 2 & -2 \\ 2 & 1 \end{bmatrix} \).
In simple words: Multiply matrix \( A \) and \( B \) to find product \( AB \). Calculate its determinant, and swap diagonal elements while changing signs of off-diagonals to get the inverse.

Exam Tip: Always state that \( |AB| \ne 0 \) to prove the inverse exists before executing the formula steps.

Question 62. Solve the following equation for x. \( \begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix} = 0 \).
Answer:
Apply the column operation \( C_1 \to C_1 + C_2 + C_3 \):
\( \begin{vmatrix} 3a-x & a-x & a-x \\ 3a-x & a+x & a-x \\ 3a-x & a-x & a+x \end{vmatrix} = 0 \).
Factor out \( (3a-x) \) from the first column:
\( (3a-x) \begin{vmatrix} 1 & a-x & a-x \\ 1 & a+x & a-x \\ 1 & a-x & a+x \end{vmatrix} = 0 \).
Apply the row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( (3a-x) \begin{vmatrix} 1 & a-x & a-x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \end{vmatrix} = 0 \).
Expanding along the first column:
\( (3a-x) \cdot 1 \cdot \left[ (2x)(2x) - 0 \right] = 0 \)
\( 4x^2 (3a-x) = 0 \).
This gives:
\( x = 0 \) or \( 3a-x = 0 \implies x = 3a \).
Thus, the solutions are \( x = 0, 3a \).
In simple words: Sum columns to find the factor \( (3a-x) \). Pull it out, use row subtraction to create zeros, expand to get a polynomial equation, and solve for \( x \).

Exam Tip: Do not divide by \( x^2 \) during calculation; doing so will lose the \( x=0 \) root, resulting in loss of marks.

Short Answer Type Questions (4 Marks)

Question 63. If \( A = \begin{bmatrix} 0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix} \) and \( I \) is the identity matrix of order 2, show that, \( I + A = (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \).
Answer:
Let \( t = \tan\frac{\alpha}{2} \).
Then, we can rewrite the matrix \( A \) as:
\( A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} \).
The left-hand side (LHS) of the equation is:
\( I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \).
Now, let us calculate the right-hand side (RHS). First, find \( I - A \):
\( I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \).
Using the half-angle trigonometric relations, we can write:
\( \cos\alpha = \frac{1 - t^2}{1 + t^2} \) and \( \sin\alpha = \frac{2t}{1 + t^2} \).
Substituting these into the given matrix:
\( \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \frac{1}{1 + t^2} \begin{bmatrix} 1 - t^2 & -2t \\ 2t & 1 - t^2 \end{bmatrix} \).
Now compute the product for RHS:
\( \text{RHS} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \left( \frac{1}{1 + t^2} \begin{bmatrix} 1 - t^2 & -2t \\ 2t & 1 - t^2 \end{bmatrix} \right) \)
\( = \frac{1}{1 + t^2} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \begin{bmatrix} 1 - t^2 & -2t \\ 2t & 1 - t^2 \end{bmatrix} \)
\( = \frac{1}{1 + t^2} \begin{bmatrix} 1(1 - t^2) + t(2t) & 1(-2t) + t(1 - t^2) \\ -t(1 - t^2) + 1(2t) & -t(-2t) + 1(1 - t^2) \end{bmatrix} \)
\( = \frac{1}{1 + t^2} \begin{bmatrix} 1 - t^2 + 2t^2 & -2t + t - t^3 \\ -t + t^3 + 2t & 2t^2 + 1 - t^2 \end{bmatrix} \)
\( = \frac{1}{1 + t^2} \begin{bmatrix} 1 + t^2 & -t(1 + t^2) \\ t(1 + t^2) & 1 + t^2 \end{bmatrix} \).
Dividing by the scalar \( (1 + t^2) \):
\( \text{RHS} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \).
Since LHS = RHS, the identity is proved.
In simple words: Represent the tangent term as \( t \), expand both sides of the equation using the half-angle trigonometric formulas, and multiply the matrices to show they are identical.

Exam Tip: Using the substitution \( t = \tan\frac{\alpha}{2} \) simplifies the algebraic multiplication immensely and prevents trig factorization errors.

Question 64. Use matrix method to solve the following system of equations : \( 5x - 7y = 2, 7x - 5y = 3 \).
Answer:
We write the system of equations in matrix format \( AX = B \):
\( A = \begin{bmatrix} 5 & -7 \\ 7 & -5 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \end{bmatrix} \), \( B = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \).
Find the determinant of \( A \):
\( |A| = 5(-5) - (-7)(7) = -25 + 49 = 24 \).
Since \( |A| \ne 0 \), a unique solution exists, given by \( X = A^{-1}B \).
Calculate the inverse matrix \( A^{-1} \):
\( A^{-1} = \frac{1}{24} \begin{bmatrix} -5 & 7 \\ -7 & 5 \end{bmatrix} \).
Now solve for \( X \):
\( X = A^{-1}B = \frac{1}{24} \begin{bmatrix} -5 & 7 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix} \)
\( = \frac{1}{24} \begin{bmatrix} -5(2) + 7(3) \\ -7(2) + 5(3) \end{bmatrix} = \frac{1}{24} \begin{bmatrix} -10 + 21 \\ -14 + 15 \end{bmatrix} = \frac{1}{24} \begin{bmatrix} 11 \\ 1 \end{bmatrix} \).
Thus, the unique solutions are \( x = \frac{11}{24} \) and \( y = \frac{1}{24} \).
In simple words: Represent the system of equations as matrix \( AX = B \). Calculate the inverse of the coefficient matrix \( A \) and multiply it by matrix \( B \) to find the values of \( x \) and \( y \).

Exam Tip: Be sure to divide each element of the resulting vector by the determinant value to state the final fractions for \( x \) and \( y \).

Long Answer Type Questions (6 Marks)

Question 65. Obtain the inverse of the following matrix using elementary row operations \( A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \).
Answer:
We write the elementary row transformation equation \( A = IA \):
\( \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \).
Swap row 1 and row 2 (\( R_1 \leftrightarrow R_2 \)):
\( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \).
Perform the row operation \( R_3 \to R_3 - 3R_1 \):
\( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{bmatrix} A \).
Perform the row operations \( R_1 \to R_1 - 2R_2 \) and \( R_3 \to R_3 + 5R_2 \):
\( \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{bmatrix} A \).
Divide row 3 by 2 (\( R_3 \to \frac{1}{2}R_3 \)):
\( \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5/2 & -3/2 & 1/2 \end{bmatrix} A \).
Perform the row operations \( R_1 \to R_1 + R_3 \) and \( R_2 \to R_2 - 2R_3 \):
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{bmatrix} A \).
Therefore, the inverse matrix is:
\( A^{-1} = \begin{bmatrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{bmatrix} \).
In simple words: Write \( A = IA \) and use row operations like swapping, multiplying, and subtracting rows to transform the left matrix into identity matrix \( I \). The right-hand side matrix becomes the inverse.

Exam Tip: Double check each row operation arithmetic on rough sheets to avoid simple fraction addition mistakes during the multi-step reduction.

Question 66. Use product \( \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \) to solve the system of equations \( x - y + 2z = 1, 2y - 3z = 1, 3x - 2y + 4z = 2 \).
Answer:
First, let us find the product of the given matrices:
\( \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} = \begin{bmatrix} 1(-2)-1(9)+2(6) & 1(0)-1(2)+2(1) & 1(1)-1(-3)+2(-2) \\ 0(-2)+2(9)-3(6) & 0(0)+2(2)-3(1) & 0(1)+2(-3)-3(-2) \\ 3(-2)-2(9)+4(6) & 3(0)-2(2)+4(1) & 3(1)-2(-3)+4(-2) \end{bmatrix} \)
\( = \begin{bmatrix} -2-9+12 & -2+2 & 1+3-4 \\ 18-18 & 4-3 & -6+6 \\ -6-18+24 & -4+4 & 3+6-8 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \).
Since the product of the two matrices is the identity matrix \( I \), they are inverses of each other.
Let \( A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \). This is the coefficient matrix of the given system of linear equations.
Thus, the inverse of \( A \) is the second matrix in the product:
\( A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \).
The system of equations can be written as \( AX = C \), where \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \) and \( C = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \).
The solution is given by:
\( X = A^{-1}C = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \)
\( = \begin{bmatrix} -2(1) + 0(1) + 1(2) \\ 9(1) + 2(1) - 3(2) \\ 6(1) + 1(1) - 2(2) \end{bmatrix} = \begin{bmatrix} -2 + 0 + 2 \\ 9 + 2 - 6 \\ 6 + 1 - 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix} \).
Therefore, the solution is \( x = 0, y = 5, z = 3 \).
In simple words: Multiply the two matrices to show their product is identity matrix \( I \). Use the second matrix as the inverse of the coefficient matrix and multiply it by the constants to solve for the variables.

Exam Tip: Never calculate the inverse from scratch when a product relation is given. Use the identity relation to save time and score full marks.

Question 67. Solve the following system of equations by matrix method, where \( x \ne 0, y \ne 0, z \ne 0 \) : \( \frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10, \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10, \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13 \).
Answer:
Let \( u = \frac{1}{x} \), \( v = \frac{1}{y} \), and \( w = \frac{1}{z} \).
The system of equations becomes:
\( 2u - 3v + 3w = 10 \)
\( u + v + w = 10 \)
\( 3u - v + 2w = 13 \).
We can write this system in the matrix form \( AU = B \):
\( A = \begin{bmatrix} 2 & -3 & 3 \\ 1 & 1 & 1 \\ 3 & -1 & 2 \end{bmatrix} \), \( U = \begin{bmatrix} u \\ v \\ w \end{bmatrix} \), \( B = \begin{bmatrix} 10 \\ 10 \\ 13 \end{bmatrix} \).
First, find the determinant of \( A \):
\( |A| = 2(2 - (-1)) - (-3)(2 - 3) + 3(-1 - 3) \)
\( = 2(3) + 3(-1) + 3(-4) = 6 - 3 - 12 = -9 \).
Since \( |A| \ne 0 \), a unique solution exists, given by \( U = A^{-1}B \).
Calculate the cofactors of the elements of \( A \):
\( A_{11} = 3 \), \( A_{12} = 1 \), \( A_{13} = -4 \)
\( A_{21} = 3 \), \( A_{22} = -5 \), \( A_{23} = -7 \)
\( A_{31} = -6 \), \( A_{32} = 1 \), \( A_{33} = 5 \).
Find the transpose of the cofactor matrix to get the adjoint:
\( \text{adj } A = \begin{bmatrix} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{bmatrix} \).
Now solve for \( U \):
\( U = A^{-1}B = \frac{1}{-9} \begin{bmatrix} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{bmatrix} \begin{bmatrix} 10 \\ 10 \\ 13 \end{bmatrix} \)
\( = -\frac{1}{9} \begin{bmatrix} 3(10) + 3(10) - 6(13) \\ 1(10) - 5(10) + 1(13) \\ -4(10) - 7(10) + 5(13) \end{bmatrix} = -\frac{1}{9} \begin{bmatrix} 30 + 30 - 78 \\ 10 - 50 + 13 \\ -40 - 70 + 65 \end{bmatrix} \)
\( = -\frac{1}{9} \begin{bmatrix} -18 \\ -27 \\ -45 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix} \).
Thus, \( u = 2 \), \( v = 3 \), and \( w = 5 \).
Substitute back to find the original variables:
\( x = \frac{1}{u} = \frac{1}{2} \)
\( y = \frac{1}{v} = \frac{1}{3} \)
\( z = \frac{1}{w} = \frac{1}{5} \).
In simple words: Substitute \( u,v,w \) for the fractions, solve the linear system using the standard matrix method, and then take the reciprocal of the solutions to find \( x,y,z \).

Exam Tip: Do not forget to take the reciprocal of your \( u, v, w \) solutions at the end to get the actual values for \( x, y, z \).

Question 68. Find \( A^{-1} \), where \( A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \), hence solve the system of linear equations : \( x + 2y - 3z = -4, 2x + 3y + 2z = 2, 3x - 3y - 4z = 11 \).
Answer:
First, find the determinant of matrix \( A \):
\( |A| = 1(3(-4) - 2(-3)) - 2(2(-4) - 2(3)) - 3(2(-3) - 3(3)) \)
\( = 1(-12 + 6) - 2(-8 - 6) - 3(-6 - 9) \)
\( = -6 + 28 + 45 = 67 \).
Since \( |A| \ne 0 \), matrix \( A \) is non-singular and invertible.
Calculate the cofactors of matrix \( A \):
\( A_{11} = -6 \), \( A_{12} = 14 \), \( A_{13} = -15 \)
\( A_{21} = 17 \), \( A_{22} = 5 \), \( A_{23} = 9 \)
\( A_{31} = 13 \), \( A_{32} = -8 \), \( A_{33} = -1 \).
Write the adjoint of \( A \):
\( \text{adj } A = \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \).
The inverse is:
\( A^{-1} = \frac{1}{67} \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \).
The system of equations has the exact coefficient matrix \( A \). Thus, we can solve using \( X = A^{-1}B \):
\( X = \frac{1}{67} \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix} \)
\( = \frac{1}{67} \begin{bmatrix} -6(-4) + 17(2) + 13(11) \\ 14(-4) + 5(2) - 8(11) \\ -15(-4) + 9(2) - 1(11) \end{bmatrix} = \frac{1}{67} \begin{bmatrix} 24 + 34 + 143 \\ -56 + 10 - 88 \\ 60 + 18 - 11 \end{bmatrix} \)
\( = \frac{1}{67} \begin{bmatrix} 201 \\ -134 \\ 67 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix} \).
Therefore, the solution is \( x = 3, y = -2, z = 1 \).
In simple words: Find the determinant and adjoint of matrix \( A \) to get \( A^{-1} \). Since the system's coefficients match matrix \( A \), multiply the inverse matrix by the constants vector to solve for \( x, y, z \).

Exam Tip: The word "hence" is extremely critical. Make sure you use the derived \( A^{-1} \) directly for the system solution step rather than trying a different system reduction technique.

Long Answer Type Questions (6 Marks)

Question 69. The sum of three numbers is 2. If we subtract the second number from twice the first number, we get 3. By adding double the second number and the third number we get 0. Represent it algebraically and find the numbers using matrix method.
Answer:
Let the three numbers be \( x \), \( y \), and \( z \).
From the given statements, we represent the algebraic equations:
1) \( x + y + z = 2 \)
2) \( 2x - y = 3 \) (which is \( 2x - y + 0z = 3 \))
3) \( 2y + z = 0 \) (which is \( 0x + 2y + z = 0 \))
We can write this system in the matrix form \( AX = B \):
\( A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 0 \\ 0 & 2 & 1 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} \).
First, find the determinant of \( A \):
\( |A| = 1(-1(1) - 0(2)) - 1(2(1) - 0(0)) + 1(2(2) - (-1)(0)) \)
\( = 1(-1) - 1(2) + 1(4) = -1 - 2 + 4 = 1 \).
Since \( |A| \ne 0 \), a unique solution exists, given by \( X = A^{-1}B \).
Now, find the cofactors of the elements of \( A \):
\( A_{11} = -1 \), \( A_{12} = -2 \), \( A_{13} = 4 \)
\( A_{21} = 1 \), \( A_{22} = 1 \), \( A_{23} = -2 \)
\( A_{31} = 1 \), \( A_{32} = 2 \), \( A_{33} = -3 \).
We write the adjoint matrix by transposing the cofactor matrix:
\( \text{adj } A = \begin{bmatrix} -1 & 1 & 1 \\ -2 & 1 & 2 \\ 4 & -2 & -3 \end{bmatrix} \).
Since \( |A| = 1 \), the inverse is:
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \begin{bmatrix} -1 & 1 & 1 \\ -2 & 1 & 2 \\ 4 & -2 & -3 \end{bmatrix} \).
Now solve for \( X \):
\( X = A^{-1}B = \begin{bmatrix} -1 & 1 & 1 \\ -2 & 1 & 2 \\ 4 & -2 & -3 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} \)
\( = \begin{bmatrix} -1(2) + 1(3) + 1(0) \\ -2(2) + 1(3) + 2(0) \\ 4(2) - 2(3) - 3(0) \end{bmatrix} = \begin{bmatrix} -2 + 3 \\ -4 + 3 \\ 8 - 6 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \).
Thus, the three numbers are \( x = 1 \), \( y = -1 \), and \( z = 2 \).
In simple words: Turn the word descriptions into three linear equations, write them in matrix form \( AX=B \), calculate the inverse of \( A \), and multiply it by the constants to find the three numbers.

Exam Tip: Be careful to include \( 0 \) coefficients for the missing terms (such as \( 0z \) in the second equation and \( 0x \) in the third) to set up your coefficient matrix properly.

Question 70. Compute the inverse of the matrix. \( A = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 5 \end{bmatrix} \) and verify that \( A^{-1} A = I_3 \).
Answer:
First, let us calculate the determinant of \( A \):
\( |A| = 3(6(5) - (-5)(-2)) - (-1)(-15(5) - (-5)(5)) + 1(-15(-2) - 6(5)) \)
\( = 3(30 - 10) + 1(-75 + 25) + 1(30 - 30) \)
\( = 3(20) - 50 + 0 = 60 - 50 = 10 \).
Since \( |A| \ne 0 \), the inverse matrix exists.
Next, compute the cofactors of \( A \):
\( A_{11} = 20 \), \( A_{12} = 50 \), \( A_{13} = 0 \)
\( A_{21} = 3 \), \( A_{22} = 10 \), \( A_{23} = 1 \)
\( A_{31} = -1 \), \( A_{32} = 0 \), \( A_{33} = 3 \).
Write the adjoint matrix:
\( \text{adj } A = \begin{bmatrix} 20 & 3 & -1 \\ 50 & 10 & 0 \\ 0 & 1 & 3 \end{bmatrix} \).
Therefore, the inverse matrix is:
\( A^{-1} = \frac{1}{10} \begin{bmatrix} 20 & 3 & -1 \\ 50 & 10 & 0 \\ 0 & 1 & 3 \end{bmatrix} \).
To verify \( A^{-1}A = I_3 \), compute the product:
\( A^{-1}A = \frac{1}{10} \begin{bmatrix} 20 & 3 & -1 \\ 50 & 10 & 0 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 5 \end{bmatrix} \)
\( = \frac{1}{10} \begin{bmatrix} 20(3)+3(-15)-1(5) & 20(-1)+3(6)-1(-2) & 20(1)+3(-5)-1(5) \\ 50(3)+10(-15)+0(5) & 50(-1)+10(6)+0(-2) & 50(1)+10(-5)+0(5) \\ 0(3)+1(-15)+3(5) & 0(-1)+1(6)+3(-2) & 0(1)+1(-5)+3(5) \end{bmatrix} \)
\( = \frac{1}{10} \begin{bmatrix} 60-45-5 & -20+18+2 & 20-15-5 \\ 150-150 & -50+60 & 50-50 \\ -15+15 & 6-6 & -5+15 \end{bmatrix} \)
\( = \frac{1}{10} \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3 \).
Thus, the relation is verified.
In simple words: Find the determinant and the cofactors of matrix \( A \). Write down \( A^{-1} \), and then multiply it by matrix \( A \) to show that the result is indeed the identity matrix.

Exam Tip: Keep the scalar fraction \( \frac{1}{10} \) outside the matrix until the end of multiplication; this keeps the elements integers and makes the matrix arithmetic much easier.

Question 71. If the matrix \( A = \begin{bmatrix} 1 & 1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \) and \( B^{-1} = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & -1 \\ 1 & 0 & 2 \end{bmatrix} \), then compute \( (AB)^{-1} \).
Answer:
Using the reversal property of inverse matrices:
\( (AB)^{-1} = B^{-1} A^{-1} \).
We are given \( B^{-1} \), so we must find \( A^{-1} \).
First, calculate the determinant of \( A \):
\( |A| = 1(2(4) - (-3)(-2)) - 1(0(4) - (-3)(3)) + 2(0(-2) - 2(3)) \)
\( = 1(8 - 6) - 1(9) + 2(-6) = 2 - 9 - 12 = -19 \).
Next, compute the cofactors of \( A \):
\( A_{11} = 2 \), \( A_{12} = -9 \), \( A_{13} = -6 \)
\( A_{21} = -8 \), \( A_{22} = -2 \), \( A_{23} = 5 \)
\( A_{31} = -7 \), \( A_{32} = 3 \), \( A_{33} = 2 \).
Write the adjoint matrix:
\( \text{adj } A = \begin{bmatrix} 2 & -8 & -7 \\ -9 & -2 & 3 \\ -6 & 5 & 2 \end{bmatrix} \).
The inverse matrix of \( A \) is:
\( A^{-1} = \frac{1}{-19} \begin{bmatrix} 2 & -8 & -7 \\ -9 & -2 & 3 \\ -6 & 5 & 2 \end{bmatrix} \).
Now, compute the product \( B^{-1}A^{-1} \):
\( (AB)^{-1} = -\frac{1}{19} \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & -1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -8 & -7 \\ -9 & -2 & 3 \\ -6 & 5 & 2 \end{bmatrix} \)
\( = -\frac{1}{19} \begin{bmatrix} 1(2)+2(-9)+0 & 1(-8)+2(-2)+0 & 1(-7)+2(3)+0 \\ 0+3(-9)-1(-6) & 0+3(-2)-1(5) & 0+3(3)-1(2) \\ 1(2)+0+2(-6) & 1(-8)+0+2(5) & 1(-7)+0+2(2) \end{bmatrix} \)
\( = -\frac{1}{19} \begin{bmatrix} -16 & -12 & -1 \\ -21 & -11 & 7 \\ -10 & 2 & -3 \end{bmatrix} = \frac{1}{19} \begin{bmatrix} 16 & 12 & 1 \\ 21 & 11 & -7 \\ 10 & -2 & 3 \end{bmatrix} \).
In simple words: Since \( (AB)^{-1} = B^{-1}A^{-1} \), calculate the inverse of \( A \) first, and then multiply the given matrix \( B^{-1} \) by the newly calculated \( A^{-1} \).

Exam Tip: Be sure to execute the multiplication in the correct order: \( B^{-1} \) on the left and \( A^{-1} \) on the right, since matrix multiplication is not commutative.

Question 72. Using matrix method, solve the following system of linear equations : \( 2x - y = 4, 2y + z = 5, z + 2x = 7 \).
Answer:
We write the given system of linear equations in matrix form \( AX = B \):
\( A = \begin{bmatrix} 2 & -1 & 0 \\ 0 & 2 & 1 \\ 2 & 0 & 1 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), \( B = \begin{bmatrix} 4 \\ 5 \\ 7 \end{bmatrix} \).
First, find the determinant of \( A \):
\( |A| = 2(2(1) - 1(0)) - (-1)(0(1) - 1(2)) + 0 = 4 - 2 = 2 \).
Since \( |A| \ne 0 \), the system has a unique solution given by \( X = A^{-1}B \).
Calculate the cofactors of the elements of \( A \):
\( A_{11} = 2 \), \( A_{12} = 2 \), \( A_{13} = -4 \)
\( A_{21} = 1 \), \( A_{22} = 2 \), \( A_{23} = -2 \)
\( A_{31} = -1 \), \( A_{32} = -2 \), \( A_{33} = 4 \).
Write the adjoint matrix:
\( \text{adj } A = \begin{bmatrix} 2 & 1 & -1 \\ 2 & 2 & -2 \\ -4 & -2 & 4 \end{bmatrix} \).
Thus, the inverse is:
\( A^{-1} = \frac{1}{2} \begin{bmatrix} 2 & 1 & -1 \\ 2 & 2 & -2 \\ -4 & -2 & 4 \end{bmatrix} \).
Now solve for \( X \):
\( X = A^{-1}B = \frac{1}{2} \begin{bmatrix} 2 & 1 & -1 \\ 2 & 2 & -2 \\ -4 & -2 & 4 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \\ 7 \end{bmatrix} \)
\( = \frac{1}{2} \begin{bmatrix} 2(4) + 1(5) - 1(7) \\ 2(4) + 2(5) - 2(7) \\ -4(4) - 2(5) + 4(7) \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 8 + 5 - 7 \\ 8 + 10 - 14 \\ -16 - 10 + 28 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 6 \\ 4 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} \).
Therefore, the solution is \( x = 3, y = 2, z = 1 \).
In simple words: Represent the system as matrix equations, calculate the determinant and inverse of the coefficients matrix \( A \), and multiply the inverse by the constants to find the values of \( x \), \( y \), and \( z \).

Exam Tip: Be sure to write the zero coefficients (like \( 0z \) in row 1, \( 0x \) in row 2, and \( 0y \) in row 3) correctly when drafting your coefficient matrix \( A \).

Question 73. Find \( A^{-1} \) if \( A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \). Also show that \( A^{-1} = \frac{A^2 - 3I}{2} \).
Answer:
First, find the determinant of \( A \):
\( |A| = 0 - 1(0 - 1) + 1(1 - 0) = 1 + 1 = 2 \).
Compute the cofactors of the elements of \( A \):
\( A_{11} = -1 \), \( A_{12} = 1 \), \( A_{13} = 1 \)
\( A_{21} = 1 \), \( A_{22} = -1 \), \( A_{23} = 1 \)
\( A_{31} = 1 \), \( A_{32} = 1 \), \( A_{33} = -1 \).
Write the adjoint matrix:
\( \text{adj } A = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \).
The inverse is:
\( A^{-1} = \frac{1}{2} \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \).
Now, calculate \( A^2 \):
\( A^2 = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0+1+1 & 0+0+1 & 0+1+0 \\ 0+0+1 & 1+0+1 & 1+0+0 \\ 0+1+0 & 1+0+0 & 1+1+0 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} \).
Compute the matrix expression \( A^2 - 3I \):
\( A^2 - 3I = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \).
Divide this expression by 2:
\( \frac{A^2 - 3I}{2} = \frac{1}{2} \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \).
Comparing this result with our derived \( A^{-1} \), we see they are completely identical.
Thus, the relation is verified.
In simple words: Find the inverse of \( A \) using determinants. Then compute the squared matrix \( A^2 \), subtract three times the identity matrix, divide by 2, and show that both resulting matrices are equal.

Exam Tip: Be sure to write out the verification steps for \( A^2 - 3I \) explicitly; do not just state the final result, as intermediate steps are graded.

Question 74. Find the inverse of the matrix \( A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \) by using elementary column transformations.
Answer:
We write the elementary column transformation equation \( A = AI \):
\( \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} = A \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
Perform the column operations \( C_2 \to C_2 - 2C_1 \) and \( C_3 \to C_3 + 2C_1 \):
\( \begin{bmatrix} 1 & 0 & 0 \\ -1 & 5 & -2 \\ 0 & -2 & 1 \end{bmatrix} = A \begin{bmatrix} 1 & -2 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
Perform the column operation \( C_2 \to C_2 + 2C_3 \):
\( \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix} = A \begin{bmatrix} 1 & 2 & 2 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{bmatrix} \).
Perform the column operations \( C_1 \to C_1 + C_2 \) and \( C_3 \to C_3 + 2C_2 \):
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = A \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \).
Therefore, the inverse matrix is:
\( A^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \).
In simple words: Write \( A = AI \), and apply column operations step-by-step to transform the left matrix into identity matrix \( I \). The right-hand side matrix becomes the inverse.

Exam Tip: For column transformations, remember you must always use the equation \( A = AI \), and execute only column-level operations; mixing row and column operations will result in zero marks.

Question 75. Let \( A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \) and \( f(x) = x^2 - 4x + 7 \). Show that f(A) = 0. Use this result to find \( A^5 \).
Answer:
First, find \( A^2 \):
\( A^2 = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4-3 & 6+6 \\ -2-2 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} \).
Compute \( f(A) \):
\( f(A) = A^2 - 4A + 7I = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).
So, \( f(A) = O \) is verified.
Using the verified relation \( A^2 - 4A + 7I = O \implies A^2 = 4A - 7I \), we express higher powers of \( A \):
\( A^3 = A \cdot A^2 = A(4A - 7I) = 4A^2 - 7A = 4(4A - 7I) - 7A = 9A - 28I \).
\( A^4 = A \cdot A^3 = A(9A - 28I) = 9A^2 - 28I \cdot A = 9(4A - 7I) - 28A = 8A - 63I \).
\( A^5 = A \cdot A^4 = A(8A - 63I) = 8A^2 - 63A = 8(4A - 7I) - 63A = -31A - 56I \).
Substitute \( A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \):
\( A^5 = -31 \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} - 56 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} -62 & -93 \\ 31 & -62 \end{bmatrix} - \begin{bmatrix} 56 & 0 \\ 0 & 56 \end{bmatrix} = \begin{bmatrix} -118 & -93 \\ 31 & -118 \end{bmatrix} \).
In simple words: Verify that substituting \( A \) into the polynomial yields zero. Use the relation \( A^2 = 4A - 7I \) recursively to express \( A^5 \) in terms of \( A \) and \( I \), and then evaluate.

Exam Tip: Using algebraic reduction to find higher matrix powers (like \( A^5 \)) is much faster and less error-prone than doing four consecutive matrix multiplications.

Question 76. If \( A = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \), verify that \( A \cdot (adj A) = (adj A) \cdot A = |A| I_3 \).
Answer:
First, find the determinant of \( A \):
\( |A| = 1(\cos^2\alpha - (-\sin^2\alpha)) = \cos^2\alpha + \sin^2\alpha = 1 \).
We need to show that both products equal \( 1 \times I_3 = I_3 \).
Calculate the cofactors of \( A \):
\( A_{11} = \cos\alpha \), \( A_{12} = -\sin\alpha \), \( A_{13} = 0 \)
\( A_{21} = \sin\alpha \), \( A_{22} = \cos\alpha \), \( A_{23} = 0 \)
\( A_{31} = 0 \), \( A_{32} = 0 \), \( A_{33} = 1 \).
Write the adjoint matrix:
\( \text{adj } A = \begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
Now compute \( A \cdot (\text{adj } A) \):
\( A \cdot (\text{adj } A) = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & \cos\alpha\sin\alpha - \sin\alpha\cos\alpha & 0 \\ \sin\alpha\cos\alpha - \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3 \).
Similarly, compute \( (\text{adj } A) \cdot A \):
\( (\text{adj } A) \cdot A = \begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & -\cos\alpha\sin\alpha + \sin\alpha\cos\alpha & 0 \\ -\sin\alpha\cos\alpha + \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3 \).
Since \( A \cdot (\text{adj } A) = (\text{adj } A) \cdot A = |A| I_3 \), the verification is complete.
In simple words: Find the determinant and cofactors of \( A \). Write down the adjoint, multiply it by matrix \( A \) in both directions, and show both products equal the identity matrix.

Exam Tip: Remember to use the key trigonometric identity \( \cos^2\alpha + \sin^2\alpha = 1 \) to simplify your diagonal elements in the final matrix multiplication.

Long Answer Type Questions (6 Marks)

Question 77. For the matrix \( A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \), verify that \( A^3 - 6A^2 + 9A - 4I = 0 \), hence find \( A^{-1} \).
Answer:
First, let us calculate \( A^2 \):
\( A^2 = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} \).
Now compute \( A^3 \):
\( A^3 = A \cdot A^2 = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} \).
Now, let us evaluate the given characteristic equation:
\( A^3 - 6A^2 + 9A - 4I = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} - \begin{bmatrix} 36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36 \end{bmatrix} + \begin{bmatrix} 18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 22-36+18-4 & -21+30-9-0 & 21-30+9-0 \\ -21+30-9-0 & 22-36+18-4 & -21+30-9-0 \\ 21-30+9-0 & -21+30-9-0 & 22-36+18-4 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O \).
Thus, the characteristic relation is verified.
To find \( A^{-1} \), we multiply both sides of the equation by \( A^{-1} \):
\( A^{-1}(A^3 - 6A^2 + 9A - 4I) = O \)
\( A^2 - 6A + 9I - 4A^{-1} = O \)
\( \implies 4A^{-1} = A^2 - 6A + 9I \).
Substitute the matrices back:
\( 4A^{-1} = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} - \begin{bmatrix} 12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \)
\( = \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix} \)
\( \implies A^{-1} = \frac{1}{4} \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix} \).
In simple words: Find the powers \( A^2 \) and \( A^3 \), substitute them into the polynomial to show it equals 0, and then rearrange the verified relation to isolate and calculate \( A^{-1} \).

Exam Tip: Never calculate \( A^{-1} \) using cofactors when the question has a "hence" instruction. You must derive the inverse directly from the verified polynomial equation.

Question 78. Find the matrix X for which \( \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix} X \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix} \).
Answer:
This matrix equation can be written as \( AXB = C \), where:
\( A = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix} \), \( B = \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} \), \( C = \begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix} \).
If both \( A \) and \( B \) are invertible, we find \( X \) as:
\( X = A^{-1} C B^{-1} \).
Calculate the inverse \( A^{-1} \):
\( |A| = 3(5) - 2(7) = 1 \ne 0 \).
\( A^{-1} = \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix} \).
Calculate the inverse \( B^{-1} \):
\( |B| = -1(1) - 1(-2) = 1 \ne 0 \).
\( B^{-1} = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \).
Now compute the product \( A^{-1}C \):
\( A^{-1}C = \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 10 & -13 \\ -14 & 19 \end{bmatrix} \).
Finally, find \( X = (A^{-1}C)B^{-1} \):
\( X = \begin{bmatrix} 10 & -13 \\ -14 & 19 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} -16 & 3 \\ 24 & -5 \end{bmatrix} \).
In simple words: Represent the equation as \( AXB = C \), calculate the inverses of matrices \( A \) and \( B \), and multiply to find the resulting matrix \( X = A^{-1}CB^{-1} \).

Exam Tip: Be highly careful with the order of operations; \( A^{-1} \) must be multiplied from the left and \( B^{-1} \) from the right, as matrix multiplication is non-commutative.

Question 79. By using properties of determinants prove the following : \( \begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} = (1+a^2+b^2)^3 \).
Answer:
Let \( \Delta \) be the determinant.
Perform the column operations \( C_1 \to C_1 - bC_3 \) and \( C_2 \to C_2 + aC_3 \):
- For column 1:
\( C_{11} = 1 + a^2 - b^2 - b(-2b) = 1 + a^2 + b^2 \)
\( C_{12} = 2ab - b(2a) = 0 \)
\( C_{13} = 2b - b(1 - a^2 - b^2) = b(1 + a^2 + b^2) \).
- For column 2:
\( C_{21} = 2ab + a(-2b) = 0 \)
\( C_{22} = 1 - a^2 + b^2 + a(2a) = 1 + a^2 + b^2 \)
\( C_{23} = -2a + a(1 - a^2 - b^2) = -a(1 + a^2 + b^2) \).
So, we can write the determinant as:
\( \Delta = \begin{vmatrix} 1+a^2+b^2 & 0 & -2b \\ 0 & 1+a^2+b^2 & 2a \\ b(1+a^2+b^2) & -a(1+a^2+b^2) & 1-a^2-b^2 \end{vmatrix} \).
Factor out \( (1 + a^2 + b^2) \) from both Column 1 and Column 2:
\( \Delta = (1 + a^2 + b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ b & -a & 1-a^2-b^2 \end{vmatrix} \).
Now perform the row operation \( R_3 \to R_3 - bR_1 + aR_2 \):
- The third element of the third row becomes: \( (1-a^2-b^2) - b(-2b) + a(2a) = 1 + a^2 + b^2 \).
The determinant reduces to:
\( \Delta = (1 + a^2 + b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ 0 & 0 & 1+a^2+b^2 \end{vmatrix} \).
Expanding along the first column yields:
\( \Delta = (1 + a^2 + b^2)^2 \cdot 1 \cdot (1 + a^2 + b^2) = (1 + a^2 + b^2)^3 \).
Hence proved.
In simple words: Subtract multiple of column 3 from the others to build common factors of \( 1+a^2+b^2 \) in columns 1 and 2, pull them out, simplify row 3 to create zeros, and expand.

Exam Tip: Generating multiple zeros in the same row or column makes the final expanding step very simple and helps avoid complex polynomial arithmetic errors.

Question 80. Prove that \( \begin{vmatrix} (y+z)^2 & xy & zx \\ xy & (x+z)^2 & yz \\ xz & yz & (x+y)^2 \end{vmatrix} = 2xyz(x+y+z)^3 \).
Answer:
Let \( \Delta \) be the given determinant.
Multiply row 1, row 2, and row 3 by \( x \), \( y \), and \( z \) respectively, and divide the determinant by \( xyz \):
\( \Delta = \frac{1}{xyz} \begin{vmatrix} x(y+z)^2 & x^2 y & x^2 z \\ xy^2 & y(x+z)^2 & y^2 z \\ xz^2 & yz^2 & z(x+y)^2 \end{vmatrix} \).
Now, factor out \( x \), \( y \), and \( z \) from columns 1, 2, and 3 respectively:
\( \Delta = \frac{xyz}{xyz} \begin{vmatrix} (y+z)^2 & x^2 & x^2 \\ y^2 & (x+z)^2 & y^2 \\ z^2 & z^2 & (x+y)^2 \end{vmatrix} = \begin{vmatrix} (y+z)^2 & x^2 & x^2 \\ y^2 & (x+z)^2 & y^2 \\ z^2 & z^2 & (x+y)^2 \end{vmatrix} \).
Apply the column operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = \begin{vmatrix} (y+z)^2 & x-y-z(x+y+z) & x-y-z(x+y+z) \\ y^2 & x+z-y(x+y+z) & 0 \\ z^2 & 0 & x+y-z(x+y+z) \end{vmatrix} \).
Factor out \( (x + y + z) \) from both Column 2 and Column 3:
\( \Delta = (x+y+z)^2 \begin{vmatrix} (y+z)^2 & x-y-z & x-y-z \\ y^2 & x+z-y & 0 \\ z^2 & 0 & x+y-z \end{vmatrix} \).
Apply the row operation \( R_1 \to R_1 - R_2 - R_3 \):
\( \Delta = (x+y+z)^2 \begin{vmatrix} 2yz & -2z & -2y \\ y^2 & x+z-y & 0 \\ z^2 & 0 & x+y-z \end{vmatrix} \).
Factor out 2 from the first row:
\( \Delta = 2(x+y+z)^2 \begin{vmatrix} yz & -z & -y \\ y^2 & x+z-y & 0 \\ z^2 & 0 & x+y-z \end{vmatrix} \).
Apply the column operations \( C_2 \to C_2 + \frac{1}{y} C_1 \) and \( C_3 \to C_3 + \frac{1}{z} C_1 \):
\( \Delta = 2(x+y+z)^2 \begin{vmatrix} yz & 0 & 0 \\ y^2 & x+z & y^2/z \\ z^2 & z^2/y & x+y \end{vmatrix} \).
Expanding along the first row:
\( \Delta = 2(x+y+z)^2 \cdot yz \left[ (x+z)(x+y) - \frac{y^2}{z} \cdot \frac{z^2}{y} \right] \)
\( = 2yz(x+y+z)^2 \left[ x^2 + xy + xz + yz - yz \right] \)
\( = 2yz(x+y+z)^2 \left[ x(x+y+z) \right] = 2xyz(x+y+z)^3 \).
Hence proved.
In simple words: Multiply rows by variables and pull them out of columns to reduce powers. Perform column differences to factor out \( (x+y+z) \), create zeros, and expand.

Exam Tip: Scaling rows and columns with variables is a key technique for handling symmetric determinants with quadratic sum entries.

Question 81. Prove that \( \begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} = a^3 \).
Answer:
Let \( \Delta \) be the determinant.
Perform row operations \( R_2 \to R_2 - 2R_1 \) and \( R_3 \to R_3 - 3R_1 \):
- For row 2:
\( R_{21} = 2a - 2a = 0 \)
\( R_{22} = 3a+2b - 2(a+b) = a \)
\( R_{23} = 4a+3b+2c - 2(a+b+c) = 2a + b \).
- For row 3:
\( R_{31} = 3a - 3a = 0 \)
\( R_{32} = 6a+3b - 3(a+b) = 3a \)
\( R_{33} = 10a+6b+3c - 3(a+b+c) = 7a + 3b \).
The determinant becomes:
\( \Delta = \begin{vmatrix} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 3a & 7a+3b \end{vmatrix} \).
Apply row operation \( R_3 \to R_3 - 3R_2 \):
- The third row elements simplify to: \( [0 \quad 0 \quad (7a+3b) - 3(2a+b)] = [0 \quad 0 \quad a] \).
So, we have:
\( \Delta = \begin{vmatrix} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 0 & a \end{vmatrix} \).
Since this is an upper triangular matrix, its determinant is simply the product of its diagonal elements:
\( \Delta = a \cdot a \cdot a = a^3 \).
Hence proved.
In simple words: Use row operations to systematically eliminate the variable terms below the main diagonal, reducing the determinant to an upper triangular form where the value is the diagonal product.

Exam Tip: Recognizing triangular structures is highly powerful; once all elements below or above the diagonal are zero, the determinant is immediately the product of diagonal elements.

Question 82. If x, y, z are different and \( \begin{vmatrix} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix} = 0 \). Show that xyz = - 1.
Answer:
We split the given determinant using the column addition property:
\( \Delta = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} = 0 \).
For the second determinant, factor out \( x \), \( y \), and \( z \) from rows 1, 2, and 3 respectively:
Second term \( = xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \).
In the first determinant, interchange column 3 with column 2, and then with column 1:
\( \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} = - \begin{vmatrix} x & 1 & x^2 \\ y & 1 & y^2 \\ z & 1 & z^2 \end{vmatrix} = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \).
Substituting these back:
\( (1 + xyz) \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = 0 \).
The standard Vandermonde determinant is:
\( \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = (x - y)(y - z)(z - x) \).
Since \( x, y, z \) are given to be different, \( (x-y) \ne 0 \), \( (y-z) \ne 0 \), and \( (z-x) \ne 0 \), meaning this determinant value cannot be zero.
Therefore, we must have:
\( 1 + xyz = 0 \implies xyz = -1 \).
Hence proved.
In simple words: Split the determinant, factor out row variables from the second part, swap columns to make the determinants match, factor out \( (1+xyz) \), and note that the remaining part is non-zero because the variables are distinct.

Exam Tip: Be sure to explicitly state that the determinant part is non-zero because \( x \ne y \ne z \) to justify the final algebraic conclusion.

Question 83. If x, y, z are the 10th, 13th and 15th terms of a G.P. find the value of \( \Delta = \begin{vmatrix} \log x & 10 & 1 \\ \log y & 13 & 1 \\ \log z & 15 & 1 \end{vmatrix} \).
Answer:
Let the first term of the G.P. be \( A \) and the common ratio be \( R \).
We can write the given terms as:
\( x = AR^9 \implies \log x = \log A + 9 \log R \)
\( y = AR^{12} \implies \log y = \log A + 12 \log R \)
\( z = AR^{14} \implies \log z = \log A + 14 \log R \).
Substitute these relations into the first column of the determinant:
\( \Delta = \begin{vmatrix} \log A + 9 \log R & 10 & 1 \\ \log A + 12 \log R & 13 & 1 \\ \log A + 14 \log R & 15 & 1 \end{vmatrix} \).
Apply the row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_2 \):
- Row 2 becomes: \( [3 \log R \quad 3 \quad 0] \)
- Row 3 becomes: \( [2 \log R \quad 2 \quad 0] \).
So, we have:
\( \Delta = \begin{vmatrix} \log A + 9 \log R & 10 & 1 \\ 3 \log R & 3 & 0 \\ 2 \log R & 2 & 0 \end{vmatrix} \).
Expanding along the third column:
\( \Delta = 1 \cdot \begin{vmatrix} 3 \log R & 3 \\ 2 \log R & 2 \end{vmatrix} = 1(6 \log R - 6 \log R) = 0 \).
In simple words: Represent \( x,y,z \) using the general term formula of G.P., apply logarithmic properties, substitute them into the matrix, create row differences, and expand to find the value of 0.

Exam Tip: Remember the logarithm property \( \log(u v) = \log u + \log v \) and \( \log(u^w) = w \log u \) to simplify geometric terms properly.

Question 84. Using the properties of determinants, show that : \( \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = abc \left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = abc + bc + ca + ab \)
Answer:
Let \( \Delta \) be the given determinant.
Factor out \( a \) from Row 1, \( b \) from Row 2, and \( c \) from Row 3:
\( \Delta = abc \begin{vmatrix} \frac{1}{a} + 1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b} + 1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c} + 1 \end{vmatrix} \).
Now perform the row transformation \( R_1 \to R_1 + R_2 + R_3 \):
\( \Delta = abc \begin{vmatrix} 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} & 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} & 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \\ \frac{1}{b} & \frac{1}{b} + 1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c} + 1 \end{vmatrix} \).
Factor out \( \left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \) from Row 1:
\( \Delta = abc \left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \begin{vmatrix} 1 & 1 & 1 \\ \frac{1}{b} & \frac{1}{b} + 1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c} + 1 \end{vmatrix} \).
Now perform the column operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = abc \left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \begin{vmatrix} 1 & 0 & 0 \\ \frac{1}{b} & 1 & 0 \\ \frac{1}{c} & 0 & 1 \end{vmatrix} \).
Expanding along the first row:
\( \Delta = abc \left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \cdot 1 \cdot (1 - 0) = abc \left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \).
Distributing \( abc \):
\( = abc + bc + ca + ab \).
Hence proved.
In simple words: Extract the factors \( a, b, c \) from the rows to form reciprocal terms. Add the rows to create a common factor across the top, pull it out, and use column operations to easily expand the remaining triangular matrix.

Exam Tip: Extracting row factors is a highly elegant and standard trick when trying to prove determinants that equal a sum of products like \( abc + bc + ca + ab \).

Question 85. Using properties of determinants prove that \( \begin{vmatrix} -bc & b^2 + bc & c^2 + bc \\ a^2 + ac & -ac & c^2 + ac \\ a^2 + ab & b^2 + ab & -ab \end{vmatrix} = (ab + bc + ca)^3 \)
Answer:
Let \( \Delta \) be the determinant.
Multiply Row 1, Row 2, and Row 3 by \( a \), \( b \), and \( c \) respectively, and divide the determinant by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} -abc & ab(b+c) & ac(b+c) \\ ab(a+c) & -abc & bc(a+c) \\ ac(a+b) & bc(a+b) & -abc \end{vmatrix} \).
Now, factor out \( a \), \( b \), and \( c \) from Column 1, Column 2, and Column 3 respectively:
\( \Delta = \frac{abc}{abc} \begin{vmatrix} -bc & a(b+c) & a(b+c) \\ b(a+c) & -ac & b(a+c) \\ c(a+b) & c(a+b) & -ab \end{vmatrix} = \begin{vmatrix} -bc & ab+ac & ab+ac \\ ab+bc & -ac & bc+ab \\ ac+bc & ac+bc & -ab \end{vmatrix} \).
Perform the row operation \( R_1 \to R_1 + R_2 + R_3 \):
- The first element of Row 1 is: \( -bc + ab + bc + ac + bc = ab + bc + ca \).
- The second element of Row 1 is: \( ab + ac - ac + ac + bc = ab + bc + ca \).
- The third element of Row 1 is: \( ab + ac + ab + bc - ab = ab + bc + ca \).
Thus, the determinant becomes:
\( \Delta = \begin{vmatrix} ab+bc+ca & ab+bc+ca & ab+bc+ca \\ b(a+c) & -ac & b(a+c) \\ c(a+b) & c(a+b) & -ab \end{vmatrix} \).
Factor out \( (ab + bc + ca) \) from the first row:
\( \Delta = (ab+bc+ca) \begin{vmatrix} 1 & 1 & 1 \\ b(a+c) & -ac & b(a+c) \\ c(a+b) & c(a+b) & -ab \end{vmatrix} \).
Perform the column operations \( C_1 \to C_1 - C_2 \) and \( C_3 \to C_3 - C_2 \):
\( \Delta = (ab+bc+ca) \begin{vmatrix} 0 & 1 & 0 \\ ab+bc+ca & -ac & ab+bc+ca \\ 0 & c(a+b) & -(ab+bc+ca) \end{vmatrix} \).
Factor out \( (ab + bc + ca) \) from both Column 1 and Column 3:
\( \Delta = (ab+bc+ca)^3 \begin{vmatrix} 0 & 1 & 0 \\ 1 & -ac & 1 \\ 0 & c(a+b) & -1 \end{vmatrix} \).
Expanding along the first row:
\( \Delta = (ab+bc+ca)^3 \left[ -1 \begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix} \right] = (ab+bc+ca)^3 \cdot [-1(-1 - 0)] = (ab+bc+ca)^3 \).
Hence proved.
In simple words: Multiply rows by variables and extract them from columns to reduce powers. Add the rows to get a common term of \( ab+bc+ca \), factor it out, use column operations to create zeros, and expand.

Exam Tip: Always look for cyclic symmetry in determinants. Row operations that add all rows together often reveal a hidden common factor.

Question 86. If \( A = \begin{bmatrix} 3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3 \end{bmatrix} \), find \( A^{-1} \) and hence solve the system of equations : \( 3x + 4y + 7z = 14 \), \( 2x - y + 3z = 4 \), \( x + 2y - 3z = 0 \).
Answer:
First, find the determinant of matrix \( A \):
\( |A| = 3(3 - 6) - 2(-12 - 14) + 1(12 - (-7)) \)
\( = 3(-3) - 2(-26) + 1(19) = -9 + 52 + 19 = 62 \).
Since \( |A| \ne 0 \), matrix \( A \) is invertible.
Calculate the cofactors of the elements of \( A \):
\( A_{11} = -3 \), \( A_{12} = 26 \), \( A_{13} = 19 \)
\( A_{21} = 9 \), \( A_{22} = -16 \), \( A_{23} = 5 \)
\( A_{31} = 5 \), \( A_{32} = -2 \), \( A_{33} = -11 \).
Write the adjoint matrix:
\( \text{adj } A = \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11 \end{bmatrix} \).
Thus, the inverse matrix is:
\( A^{-1} = \frac{1}{62} \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11 \end{bmatrix} \).
Now consider the given system of linear equations:
\( 3x + 4y + 7z = 14 \)
\( 2x - y + 3z = 4 \)
\( x + 2y - 3z = 0 \).
The coefficient matrix of this system is \( A^T \) (the transpose of matrix \( A \)):
\( A^T = \begin{bmatrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{bmatrix} \).
We can represent this system as \( A^T X = B \), where \( B = \begin{bmatrix} 14 \\ 4 \\ 0 \end{bmatrix} \).
Using the identity \( (A^T)^{-1} = (A^{-1})^T \), the solution is given by:
\( X = (A^{-1})^T B = \frac{1}{62} \begin{bmatrix} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{bmatrix} \begin{bmatrix} 14 \\ 4 \\ 0 \end{bmatrix} \)
\( = \frac{1}{62} \begin{bmatrix} -3(14) + 26(4) + 19(0) \\ 9(14) - 16(4) + 5(0) \\ 5(14) - 2(4) - 11(0) \end{bmatrix} = \frac{1}{62} \begin{bmatrix} -42 + 104 \\ 126 - 64 \\ 70 - 8 \end{bmatrix} = \frac{1}{62} \begin{bmatrix} 62 \\ 62 \\ 62 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \).
Therefore, the solution is \( x = 1 \), \( y = 1 \), and \( z = 1 \).
In simple words: Find the inverse of matrix \( A \) first. Since the equations use the transpose of \( A \) as their coefficient matrix, transpose \( A^{-1} \) and multiply it by the constant numbers to find the values of \( x, y, z \).

Exam Tip: Be highly careful when setting up the system of equations. Since the coefficient matrix is the transpose of \( A \), you must transpose your calculated \( A^{-1} \) before multiplying by the constants matrix.