CBSE Class 12 Mathematics Determinants Worksheet Set 04

CBSE Class 12 Mathematics Determinants Worksheet (4). The Determinants questions in the worksheets have been specifically designed by best mathematics teachers so that the students can practise them to clear their Determinants concepts and get better marks in class 12 mathematics tests and examinations. Students can free download these Determinants worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Determinants chapter and other subjects too. Use them for better understanding of the subjects.

Question 1. Show that \( \begin{vmatrix} 1 & 1 & 1 \\ ^{m}\text{C}_{1} & ^{m+1}\text{C}_{1} & ^{m+2}\text{C}_{1} \\ ^{m}\text{C}_{2} & ^{m+1}\text{C}_{2} & ^{m+2}\text{C}_{2} \end{vmatrix} = 1 \).
Answer: Let us write down the given determinant on the left-hand side (L.H.S.):
\( \Delta = \begin{vmatrix} 1 & 1 & 1 \\ ^{m}\text{C}_{1} & ^{m+1}\text{C}_{1} & ^{m+2}\text{C}_{1} \\ ^{m}\text{C}_{2} & ^{m+1}\text{C}_{2} & ^{m+2}\text{C}_{2} \end{vmatrix} \)
Using the standard combination formulas \( ^{n}\text{C}_{1} = n \) and \( ^{n}\text{C}_{2} = \frac{n(n-1)}{2} \), we substitute the corresponding values into the matrix:
\( \Delta = \begin{vmatrix} 1 & 1 & 1 \\ m & m + 1 & m + 2 \\ \frac{m(m-1)}{2} & \frac{(m+1)m}{2} & \frac{(m+2)(m+1)}{2} \end{vmatrix} \)
Now, we factor out \( \frac{1}{2} \) from the third row \( R_3 \):
\( \Delta = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ m & m + 1 & m + 2 \\ m^2 - m & m^2 + m & m^2 + 3m + 2 \end{vmatrix} \)
We apply the column transformations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = \frac{1}{2} \begin{vmatrix} 1 & 0 & 0 \\ m & 1 & 2 \\ m^2 - m & 2m & 4m + 2 \end{vmatrix} \)
Expanding this determinant along the first row \( R_1 \):
\( \Delta = \frac{1}{2} [1 \cdot (1(4m + 2) - 2(2m))] \)

\( \implies \Delta = \frac{1}{2} [4m + 2 - 4m] \)

\( \implies \Delta = \frac{1}{2} [2] = 1 = \text{R.H.S.} \)
Thus, the equation is verified.
In simple words: Simplify the binomial combinations in the rows using standard formulas. Pull out the common fraction from the bottom row, perform column subtractions to create zeros, and expand to find that the value is 1.

Exam Tip: Be sure to write the combination identities \( ^{n}\text{C}_{1} = n \) and \( ^{n}\text{C}_{2} = \frac{n(n-1)}{2} \) on the side to earn full marks for explanation steps.

Question 2. Show that \( \begin{vmatrix} (b + c)^2 & ba & ca \\ ab & (c + a)^2 & cb \\ ac & bc & (a + b)^2 \end{vmatrix} = 2abc(a + b + c)^3 \).
Answer: Let us begin with the left-hand side (L.H.S.) of the determinant equation:
\( \Delta = \begin{vmatrix} (b + c)^2 & ba & ca \\ ab & (c + a)^2 & cb \\ ac & bc & (a + b)^2 \end{vmatrix} \)
We multiply the rows \( R_1, R_2, R_3 \) by \( a, b, c \) respectively. To balance this operation, we divide the entire determinant by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} a(b + c)^2 & ba^2 & ca^2 \\ ab^2 & b(c + a)^2 & cb^2 \\ ac^2 & bc^2 & c(a + b)^2 \end{vmatrix} \)
Now, we take \( a, b, c \) as common factors from the columns \( C_1, C_2, C_3 \) respectively:
\( \Delta = \frac{abc}{abc} \begin{vmatrix} (b + c)^2 & a^2 & a^2 \\ b^2 & (c + a)^2 & b^2 \\ c^2 & c^2 & (a + b)^2 \end{vmatrix} = \begin{vmatrix} (b + c)^2 & a^2 & a^2 \\ b^2 & (c + a)^2 & b^2 \\ c^2 & c^2 & (a + b)^2 \end{vmatrix} \)
Next, we apply the column operations \( C_1 \to C_1 - C_3 \) and \( C_2 \to C_2 - C_3 \):
\( \Delta = \begin{vmatrix} (b + c + a)(b + c - a) & 0 & a^2 \\ 0 & (c + a + b)(c + a - b) & b^2 \\ (c + a + b)(c - a - b) & (c + a + b)(c - a - b) & (a + b)^2 \end{vmatrix} \)
We take \( (a + b + c) \) as a common factor from both \( C_1 \) and \( C_2 \):
\( \Delta = (a + b + c)^2 \begin{vmatrix} b + c - a & 0 & a^2 \\ 0 & c + a - b & b^2 \\ c - a - b & c - a - b & (a + b)^2 \end{vmatrix} \)
We now apply the row operation \( R_3 \to R_3 - (R_1 + R_2) \):
\( \Delta = (a + b + c)^2 \begin{vmatrix} b + c - a & 0 & a^2 \\ 0 & c + a - b & b^2 \\ -2b & -2a & 2ab \end{vmatrix} \)
To simplify further, we perform \( C_1 \to aC_1 \) and \( C_2 \to bC_2 \), which requires dividing the determinant by \( ab \):
\( \Delta = \frac{(a + b + c)^2}{ab} \begin{vmatrix} ab + ac - a^2 & 0 & a^2 \\ 0 & bc + ab - b^2 & b^2 \\ -2ab & -2ab & 2ab \end{vmatrix} \)
Now, apply \( C_1 \to C_1 + C_3 \) and \( C_2 \to C_2 + C_3 \):
\( \Delta = \frac{(a + b + c)^2}{ab} \begin{vmatrix} ab + ac & a^2 & a^2 \\ b^2 & bc + ab & b^2 \\ 0 & 0 & 2ab \end{vmatrix} \)
We take the common factors \( a \) from \( R_1 \), \( b \) from \( R_2 \), and \( 2ab \) from \( R_3 \):
\( \Delta = \frac{(a + b + c)^2}{ab} \cdot ab(2ab) \begin{vmatrix} b + c & a & a \\ b & c + a & b \\ 0 & 0 & 1 \end{vmatrix} \)

\( \implies \Delta = 2ab(a + b + c)^2 \begin{vmatrix} b + c & a & a \\ b & c + a & b \\ 0 & 0 & 1 \end{vmatrix} \)
Expanding along the third row \( R_3 \):
\( \Delta = 2ab(a + b + c)^2 [1 \cdot ((b + c)(c + a) - ab)] \)

\( \implies \Delta = 2ab(a + b + c)^2 (bc + ab + c^2 + ac - ab) \)

\( \implies \Delta = 2ab(a + b + c)^2 \cdot c(b + c + a) \)

\( \implies \Delta = 2abc(a + b + c)^3 = \text{R.H.S.} \)
The identity is verified.
In simple words: Multiply rows by factors, divide the outside, and then factor the columns to change the matrix variables. Use subtraction operations to reduce the numbers to simpler terms, and expand along the easiest row.

Exam Tip: Whenever multiplying rows by variables like \( a, b, c \), always remember to divide the determinant on the outside by the product \( abc \) to keep the value equivalent.

Question 3. Show that \( \begin{vmatrix} 1 + a^2 - b^2 & 2ab & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} = (1 + a^2 + b^2)^3 \).
Answer: Let us denote the determinant as:
\( \Delta = \begin{vmatrix} 1 + a^2 - b^2 & 2ab & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} \)
We perform the key column transformations \( C_1 \to C_1 - bC_3 \) and \( C_2 \to C_2 + aC_3 \):
\( \Delta = \begin{vmatrix} 1 + a^2 - b^2 + 2b^2 & 2ab - 2ab & -2b \\ 2ab - 2ab & 1 - a^2 + b^2 + 2a^2 & 2a \\ 2b - b(1 - a^2 - b^2) & -2a + a(1 - a^2 - b^2) & 1 - a^2 - b^2 \end{vmatrix} \)
Simplifying the terms inside the matrix yields:
\( \Delta = \begin{vmatrix} 1 + a^2 + b^2 & 0 & -2b \\ 0 & 1 + a^2 + b^2 & 2a \\ b(1 + a^2 + b^2) & -a(1 + a^2 + b^2) & 1 - a^2 - b^2 \end{vmatrix} \)
We take the common factor \( (1 + a^2 + b^2) \) out from both \( C_1 \) and \( C_2 \):
\( \Delta = (1 + a^2 + b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ b & -a & 1 - a^2 - b^2 \end{vmatrix} \)
Now, we expand the determinant along the first row \( R_1 \):
\( \Delta = (1 + a^2 + b^2)^2 [1 \cdot (1(1 - a^2 - b^2) - 2a(-a)) - 2b(0 - b)] \)

\( \implies \Delta = (1 + a^2 + b^2)^2 [1 - a^2 - b^2 + 2a^2 + 2b^2] \)

\( \implies \Delta = (1 + a^2 + b^2)^2 (1 + a^2 + b^2) \)

\( \implies \Delta = (1 + a^2 + b^2)^3 = \text{R.H.S.} \)
This completes the proof.
In simple words: Add multiples of the third column to the first and second columns to create a common factor. Pull that factor out of both columns, then expand the remaining simpler matrix to find the final value.

Exam Tip: Recognizing the starting step of \( C_1 \to C_1 - bC_3 \) is key here. Practice identifying which column operations will generate identical terms across columns.

Solving System of Linear Equations (Matrix Method)

Question 4. Solve the equations using matrix method \( x + 2y + z = 7 \); \( x + 3z = 11 \); \( 2x - 3y = 1 \).
Answer: We write the system of equations with placeholders for missing variables:
\( x + 2y + z = 7 \)
\( x + 0y + 3z = 11 \)
\( 2x - 3y + 0z = 1 \)
We represent this system in matrix form \( AX = B \):
\( \begin{bmatrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 7 \\ 11 \\ 1 \end{bmatrix} \)
Here, the coefficient matrix is \( A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix} \). Let us determine its determinant:
\( |A| = 1(0 - (-9)) - 2(0 - 6) + 1(-3 - 0) \)

\( \implies |A| = 9 + 12 - 3 = 18 \)
Since \( |A| \neq 0 \), the system is consistent and has a unique solution given by \( X = A^{-1}B \).
Next, we compute the cofactors of the elements of \( A \):
\( c_{11} = 9, \quad c_{12} = 6, \quad c_{13} = -3 \)
\( c_{21} = -3, \quad c_{22} = -2, \quad c_{23} = 7 \)
\( c_{31} = 6, \quad c_{32} = -2, \quad c_{33} = -2 \)
Thus, the adjoint matrix of \( A \) is:
\( adj(A) = \begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix} \)
We find the inverse matrix \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{18} \begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix} \)
Now, we calculate the solution vector \( X \):
\( X = A^{-1}B = \frac{1}{18} \begin{bmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{bmatrix} \begin{bmatrix} 7 \\ 11 \\ 1 \end{bmatrix} \)

\( \implies X = \frac{1}{18} \begin{bmatrix} 9(7) - 3(11) + 6(1) \\ 6(7) - 2(11) - 2(1) \\ -3(7) + 7(11) - 2(1) \end{bmatrix} \)

\( \implies X = \frac{1}{18} \begin{bmatrix} 63 - 33 + 6 \\ 42 - 22 - 2 \\ -21 + 77 - 2 \end{bmatrix} = \frac{1}{18} \begin{bmatrix} 36 \\ 18 \\ 54 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \)
Therefore, the unique solution is \( x = 2 \), \( y = 1 \), and \( z = 3 \).
In simple words: Convert the system into the matrix equation \( AX = B \). Find the determinant of \( A \) to ensure a solution exists, calculate its cofactors to find the inverse, and then multiply this inverse by matrix \( B \) to find the variables.

Exam Tip: Always double check your calculated values of \( x, y, z \) by substituting them back into any of the original equations to confirm they are correct.

Question 5. Solve the equations \( \frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10 \); \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10 \); \( \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13 \).
Answer: Let us substitute new variables for the reciprocals:
Let \( u = \frac{1}{x} \), \( v = \frac{1}{y} \), and \( w = \frac{1}{z} \).
The given system transforms into the following linear equations:
\( 2u - 3v + 3w = 10 \)
\( u + v + w = 10 \)
\( 3u - v + 2w = 13 \)
We express these equations in matrix form as \( AX = B \):
\( \begin{bmatrix} 2 & -3 & 3 \\ 1 & 1 & 1 \\ 3 & -1 & 2 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 10 \\ 10 \\ 13 \end{bmatrix} \)
Here, the coefficient matrix is \( A = \begin{bmatrix} 2 & -3 & 3 \\ 1 & 1 & 1 \\ 3 & -1 & 2 \end{bmatrix} \). Let us determine its determinant:
\( |A| = 2(2 - (-1)) - (-3)(2 - 3) + 3(-1 - 3) \)

\( \implies |A| = 6 - 3 - 12 = -9 \)
As \( |A| \neq 0 \), a unique solution exists. Next, we determine the cofactors of the elements of \( A \):
\( c_{11} = 3, \quad c_{12} = 1, \quad c_{13} = -4 \)
\( c_{21} = 3, \quad c_{22} = -5, \quad c_{23} = -7 \)
\( c_{31} = -6, \quad c_{32} = 1, \quad c_{33} = 5 \)
The adjoint matrix is:
\( adj(A) = \begin{bmatrix} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{bmatrix} \)
The inverse matrix \( A^{-1} \) is:
\( A^{-1} = -\frac{1}{9} \begin{bmatrix} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{bmatrix} \)
Using \( X = A^{-1}B \), we calculate the values of \( u, v, w \):
\( X = -\frac{1}{9} \begin{bmatrix} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{bmatrix} \begin{bmatrix} 10 \\ 10 \\ 13 \end{bmatrix} \)

\( \implies X = -\frac{1}{9} \begin{bmatrix} 3(10) + 3(10) - 6(13) \\ 1(10) - 5(10) + 1(13) \\ -4(10) - 7(10) + 5(13) \end{bmatrix} \)

\( \implies X = -\frac{1}{9} \begin{bmatrix} 30 + 30 - 78 \\ 10 - 50 + 13 \\ -40 - 70 + 65 \end{bmatrix} = -\frac{1}{9} \begin{bmatrix} -18 \\ -27 \\ -45 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix} \)
Thus, we have:
\( u = 2 \implies x = \frac{1}{2} \)
\( v = 3 \implies y = \frac{1}{3} \)
\( w = 5 \implies z = \frac{1}{5} \)
The final solution is \( x = \frac{1}{2} \), \( y = \frac{1}{3} \), and \( z = \frac{1}{5} \).
In simple words: Substitute \( u, v, w \) for the fraction terms to get normal equations. Solve for \( u, v, w \) by matrix inversion, and then invert the values back to find \( x, y, z \).

Exam Tip: Do not forget the final step of taking the reciprocal of \( u, v, w \) to find \( x, y, z \), as students often stop after solving for \( u, v, w \).

Question 6. Find \( A^{-1} \), where \( A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \). Hence solve the system of equations \( x + 2y - 3z = -4 \), \( 2x + 3y + 2z = 2 \) and \( 3x - 3y - 4z = 11 \).
Answer: We are given the matrix:
\( A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \)
First, we find the determinant of \( A \):
\( |A| = 1(3(-4) - 2(-3)) - 2(2(-4) - 2(3)) - 3(2(-3) - 3(3)) \)

\( \implies |A| = 1(-12 + 6) - 2(-8 - 6) - 3(-6 - 9) \)

\( \implies |A| = -6 + 28 + 45 = 67 \)
Since \( |A| \neq 0 \), \( A \) is invertible and the system of equations has a unique solution.
Now, we compute the cofactors of matrix \( A \):
\( c_{11} = -6, \quad c_{12} = 14, \quad c_{13} = -15 \)
\( c_{21} = 17, \quad c_{22} = 5, \quad c_{23} = 9 \)
\( c_{31} = 13, \quad c_{32} = -8, \quad c_{33} = -1 \)
Thus, the adjoint of \( A \) is:
\( adj(A) = \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \)
The inverse matrix \( A^{-1} \) is:
\( A^{-1} = \frac{1}{67} \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \quad \text{- (1)} \)
Next, we write the system of linear equations in matrix form \( AX = B \):
\( \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix} \)
The solution of this system is given by \( X = A^{-1}B \). Substituting \( A^{-1} \) from equation (1):
\( X = \frac{1}{67} \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix} \)

\( \implies X = \frac{1}{67} \begin{bmatrix} -6(-4) + 17(2) + 13(11) \\ 14(-4) + 5(2) - 8(11) \\ -15(-4) + 9(2) - 1(11) \end{bmatrix} \)

\( \implies X = \frac{1}{67} \begin{bmatrix} 24 + 34 + 143 \\ -56 + 10 - 88 \\ 60 + 18 - 11 \end{bmatrix} = \frac{1}{67} \begin{bmatrix} 201 \\ -134 \\ 67 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix} \)
Therefore, the required solution is \( x = 3 \), \( y = -2 \), and \( z = 1 \).
In simple words: First calculate the inverse of the given matrix \( A \). Notice that the coefficient matrix of the system of equations is exactly \( A \), and multiply its inverse by the constants matrix on the right to solve for the variables.

Exam Tip: Look carefully at the coefficients in the system of equations to ensure they match matrix \( A \). If they do, you can use the calculated \( A^{-1} \) directly without repeating any steps.

Question 7. If A = \( \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \). Find \( A^{-1} \) and hence solve the equation \( x + 2y + z = 4 \); \( -x + y + z = 0 \) and \( x - 3y + z = 2 \).
Answer: We are given the matrix:
\( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \)
First, we calculate the determinant of \( A \):
\( |A| = 1(1 - (-3)) - (-1)(2 - (-3)) + 1(2 - 1) = 4 + 5 + 1 = 10 \)
Since \( |A| \neq 0 \), \( A \) is invertible. The cofactor matrix of \( A \) is determined to find the adjoint:
\( adj(A) = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \)
Thus, the inverse of \( A \) is:
\( A^{-1} = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \quad \text{- (1)} \)
Now, the given system of linear equations is:
\( x + 2y + z = 4 \)
\( -x + y + z = 0 \)
\( x - 3y + z = 2 \)
We can write this in matrix form as \( A'X = B \), where \( A' \) is the transpose of matrix \( A \):
\( A' = \begin{bmatrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \text{and} \quad B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} \)
To solve for \( X \), we use the property \( (A')^{-1} = (A^{-1})' \):
\( X = (A')^{-1}B = (A^{-1})'B \)
Substituting the transpose of \( A^{-1} \) from equation (1):
\( X = \frac{1}{10} \begin{bmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} \)

\( \implies X = \frac{1}{10} \begin{bmatrix} 4(4) - 5(0) + 1(2) \\ 2(4) + 0(0) - 2(2) \\ 2(4) + 5(0) + 3(2) \end{bmatrix} \)

\( \implies X = \frac{1}{10} \begin{bmatrix} 16 + 2 \\ 8 - 4 \\ 8 + 6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 18 \\ 4 \\ 14 \end{bmatrix} = \begin{bmatrix} 9/5 \\ 2/5 \\ 7/5 \end{bmatrix} \)
Thus, the solution is \( x = \frac{9}{5} \), \( y = \frac{2}{5} \), and \( z = \frac{7}{5} \).
In simple words: Find the inverse of matrix \( A \). The coefficient matrix of the system is the transpose of \( A \). To solve the system, multiply the transpose of your calculated \( A^{-1} \) by the constants vector.

Exam Tip: Notice that the coefficient matrix is the transpose of the given matrix \( A \). Use the transpose of \( A^{-1} \) to solve it, rather than calculating a new inverse matrix from scratch.

Question 8. Determine the product \( \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \) and hence (or) use it to solve the equations \( x - y + z = 4 \); \( x - 2y - 2z = 9 \); \( 2x + y + 3z = 1 \).
Answer: Let us define the two given matrices as:
\( C = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \quad \text{and} \quad A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \)
Now, we calculate their product \( CA \):
\( CA = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \)

\( \implies CA = \begin{bmatrix} -4(1)+4(1)+4(2) & -4(-1)+4(-2)+4(1) & -4(1)+4(-2)+4(3) \\ -7(1)+1(1)+3(2) & -7(-1)+1(-2)+3(1) & -7(1)+1(-2)+3(3) \\ 5(1)-3(1)-1(2) & 5(-1)-3(-2)-1(1) & 5(1)-3(-2)-1(3) \end{bmatrix} \)

\( \implies CA = \begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix} = 8I \)
Since \( CA = 8I \), we have:
\( A^{-1} = \frac{1}{8}C = \frac{1}{8} \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \)
The system of equations can be written in matrix format as \( AX = B \):
\( \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix} \)
The solution is \( X = A^{-1}B \). Substituting our inverse matrix:
\( X = \frac{1}{8} \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix} \)

\( \implies X = \frac{1}{8} \begin{bmatrix} -4(4) + 4(9) + 4(1) \\ -7(4) + 1(9) + 3(1) \\ 5(4) - 3(9) - 1(1) \end{bmatrix} \)

\( \implies X = \frac{1}{8} \begin{bmatrix} -16 + 36 + 4 \\ -28 + 9 + 3 \\ 20 - 27 - 1 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 24 \\ -16 \\ -8 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \\ -1 \end{bmatrix} \)
Thus, the solution is \( x = 3 \), \( y = -2 \), and \( z = -1 \).
In simple words: Multiply the two matrices to find that their product is \( 8I \). This tells us the inverse is \( \frac{1}{8} \) of the first matrix. Multiply this inverse by the constant values of the equations to solve for \( x \), \( y \), and \( z \).

Exam Tip: If the product of two matrices \( CA \) is a scalar multiple of the identity matrix \( kI \), you can directly find the inverse using \( A^{-1} = \frac{1}{k}C \) without doing cofactor calculations.

Question 9. A = \( \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} \) and B = \( \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \) find AB and hence solve the equations \( x - 2y = 10 \); \( 2x + y + 3z = 8 \); \( -2y + z = 7 \).
Answer: We begin by determining the product of matrices \( A \) and \( B \):
\( AB = \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \)
Multiplying the elements:
\( AB = \begin{bmatrix} 1(7) - 2(-2) + 0 & 1(2) - 2(1) + 0 & 1(-6) - 2(-3) + 0 \\ 2(7) + 1(-2) + 3(-4) & 2(2) + 1(1) + 3(2) & 2(-6) + 1(-3) + 3(5) \\ 0 - 2(-2) + 1(-4) & 0 - 2(1) + 1(2) & 0 - 2(-3) + 1(5) \end{bmatrix} \)

\( \implies AB = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = 11I \)
Using this result, we pre-multiply both sides by \( A^{-1} \) to obtain:
\( A^{-1}(AB) = 11A^{-1}I \)

\( \implies B = 11A^{-1} \)

\( \implies A^{-1} = \frac{1}{11}B = \frac{1}{11} \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \)
Now, the system of linear equations is:
\( x - 2y + 0z = 10 \)
\( 2x + y + 3z = 8 \)
\( 0x - 2y + z = 7 \)
We represent these equations in matrix format as \( AX = C \):
\( \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \)
To find the variables, we calculate \( X = A^{-1}C \):
\( X = \frac{1}{11} \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \)

\( \implies X = \frac{1}{11} \begin{bmatrix} 70 + 16 - 42 \\ -20 + 8 - 21 \\ -40 + 16 + 35 \end{bmatrix} \)

\( \implies X = \frac{1}{11} \begin{bmatrix} 44 \\ -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 4 \\ -3 \\ 1 \end{bmatrix} \)
Thus, the solution is \( x = 4 \), \( y = -3 \), and \( z = 1 \).
In simple words: Multiply the matrices to see that \( AB = 11I \), which gives the inverse of \( A \). Set up the system of equations as \( AX = C \), and multiply \( A^{-1} \) by the matrix of constants to find the solution.

Exam Tip: Always relate the system of equations back to the product computed in the first part. You can directly write \( A^{-1} \) as \( \frac{1}{11}B \), saving valuable exam time.

Question 10. Show that system of equations is consistent and also find the solution \( 2x - y + 3z = 5 \); \( 3x + 2y - z = 7 \); \( 4x + 5y - 5z = 9 \).
Answer: We write the system of equations in matrix form \( AX = B \):
\( \begin{bmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 - 5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ 7 \\ 9 \end{bmatrix} \)
Let us determine the determinant of the coefficient matrix \( A \):
\( |A| = 2(-10 - (-5)) - (-1)(-15 - (-4)) + 3(15 - 8) \)

\( \implies |A| = 2(-5) + 1(-11) + 3(7) = -10 - 11 + 21 = 0 \)
Since the determinant is zero, the system may have infinitely many solutions or no solution. We calculate \( (adj(A))B \) to check:
The adjoint of \( A \) is computed using its cofactors:
\( adj(A) = \begin{bmatrix} -5 & 10 & -5 \\ 11 & -22 & 11 \\ 7 & -14 & 7 \end{bmatrix} \)
Now, we evaluate the product \( (adj(A))B \):
\( (adj(A))B = \begin{bmatrix} -5 & 10 & -5 \\ 11 & -22 & 11 \\ 7 & -14 & 7 \end{bmatrix} \begin{bmatrix} 5 \\ 7 \\ 9 \end{bmatrix} = \begin{bmatrix} -25 + 70 - 45 \\ 55 - 154 + 99 \\ 35 - 98 + 63 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = O \)
Because \( |A| = 0 \) and \( (adj(A))B = O \), the system is consistent and possesses infinitely many solutions.
To find the parametric solutions, let \( z = k \) (where \( k \in \mathbb{R} \)). Substituting \( z = k \) into the first two equations:
\( 2x - y = 5 - 3k \)
\( 3x + 2y = 7 + k \)
We rewrite this sub-system as \( A_1 X_1 = B_1 \):
\( \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 - 3k \\ 7 + k \end{bmatrix} \)
Here, the determinant is \( |A_1| = 4 - (-3) = 7 \), and the adjoint is:
\( adj(A_1) = \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} \)
Thus, the inverse is \( A_1^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} \).
Solving for \( x \) and \( y \):
\( \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 5 - 3k \\ 7 + k \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 2(5 - 3k) + 1(7 + k) \\ -3(5 - 3k) + 2(7 + k) \end{bmatrix} \)

\( \implies \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 17 - 5k \\ 11k - 1 \end{bmatrix} \)
Therefore, the infinite set of solutions is \( x = \frac{17 - 5k}{7} \), \( y = \frac{11k - 1}{7} \), and \( z = k \), where \( k \) is any real number.
In simple words: When the determinant of the coefficient matrix is zero and \( (adj(A))B \) is also zero, the system has infinitely many solutions. Solve by substituting \( z = k \), reducing the system to two equations, and expressing \( x \) and \( y \) in terms of \( k \).

Exam Tip: Clearly state the condition \( |A| = 0 \) and \( (adj A)B = O \) to justify why you are introducing a variable parameter \( k \).

Please click the link below to download CBSE Class 12 Mathematics Determinants Worksheet (4).