ML Aggarwal Class 12 Maths Solutions Section A Chapter 02 Matrices

Introduction

Matrices came into existence through various linear problems, the most significant being the study of how systems of simultaneous linear equations work. Arthur Cayley created the foundational ideas of matrices as a mathematical system in 1857. Today, matrices play an important role in many fields including Algebra, Geometry, Statistics, Physics, Chemistry, and Psychology. This chapter focuses on learning about matrices, examining the basic rules of matrix algebra, and using these rules to solve systems of simultaneous linear equations with 2 or 3 unknowns.

2.1 Matrices

Consider the systems of simultaneous linear equations:

2x - 3y = 7
5x + 4y = 11 ... (i)

This is a system with two linear equations involving two unknowns, x and y. Also consider:

2x - 7y + 3z = 8
3x + 0y + 6z = -11 ... (ii)

This is a system with two linear equations involving three unknowns, x, y, z. In the first case, to understand the left side, we only need the collection
\( \begin{bmatrix} 2 & -3 \\ 5 & 4 \end{bmatrix} \) which holds the coefficients of x and y, in that exact order. Similarly, the collection
\( \begin{bmatrix} 7 \\ 11 \end{bmatrix} \) shows all information about the right side. Thus, the system of equations in (i) is fully specified by the two collections
\( \begin{bmatrix} 2 & -3 \\ 5 & 4 \end{bmatrix} \) and \( \begin{bmatrix} 7 \\ 11 \end{bmatrix} \).

Similarly, the system of equations in (ii) is fully specified by the two collections
\( \begin{bmatrix} 2 & -7 & 3 \\ 3 & 0 & 6 \end{bmatrix} \) and \( \begin{bmatrix} 8 \\ -11 \end{bmatrix} \).

Each collection of this type is known as a matrix. The numbers that make up each matrix are called the elements or entries. In the examples above, all elements are real numbers. In fact, the elements can be complex numbers, polynomials, or even more general functions.

For each element of a matrix, both its value and its position matter. This means that a matrix is an ordered array of elements.

In the (i) system of equations, the collection [2 -3] specifies the left side of the first equation and [5 4] specifies the left side of the second equation. Each horizontal collection of a matrix is called a row of the matrix.

Similarly, the collection
\( \begin{bmatrix} 2 \\ 5 \end{bmatrix} \) shows the coefficients of x and the collection
\( \begin{bmatrix} -3 \\ 4 \end{bmatrix} \) shows the coefficients of y in the (i) system. Each vertical line is called a column of the matrix.

Thus, the matrix
\( \begin{bmatrix} 2 & -3 \\ 5 & 4 \end{bmatrix} \) has two rows and two columns, or briefly is a 2 × 2 matrix, while the matrix
\( \begin{bmatrix} 2 & -7 & 3 \\ 3 & 0 & 6 \end{bmatrix} \) corresponding to the (ii) system is a matrix with two rows and three columns, or briefly a 2 × 3 matrix. The matrices for the right side of the above systems (i) and (ii) are 2 × 1 matrices.

Matrix Definition. A rectangular arrangement of mn elements in the form of an ordered set of m rows, each row containing an ordered set of n elements, is called an m × n matrix (m × n is read as m by n). Such an arrangement is placed in brackets [ ] or parenthesis ( ). Each of the mn elements making up the matrix is called an element or entry of the matrix. To find where a specific element is located in a matrix, we must state the row number and the column number in which the element appears.

An element in the ith row and the jth column is called the (i, j)th element or (i, j)th entry. Usually, a matrix is shown using a capital letter, and an element of a matrix using a small letter along with two subscripts - the first showing the row number and the second showing the column number in which the element appears. Thus, an m × n matrix may be written as:

A = \( \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & a_{m3} & \cdots & a_{mn} \end{bmatrix} \)

In compact form, this matrix is written as A = [aij]m × n, where
\( 1 \leq i \leq m, 1 \leq j \leq n \).

A is called a matrix of order (or type) m × n.

If aij ∈ R (the set of real numbers), we say that A is a matrix over R or A is a real matrix. Similarly, if aij ∈ C (complex numbers), then A is a complex matrix.

The collection [ai1 ai2 ai3 ... ain] is the ith row and the collection
\( \begin{bmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{mj} \end{bmatrix} \) is the jth column of the matrix A.

Comparable Matrices. Two matrices are said to be comparable if each one has the same number of rows and columns as the other.

Equal Matrices. Two matrices A = [aij] and B = [bij] are said to be equal (written as A = B) if:

(i) they have the same order - that is, the number of rows in A equals the number of rows in B and the number of columns in A equals the number of columns in B, and

(ii) their matching elements are equal - that is, aij = bij for all i and j.

For example, the matrices
\( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) and \( \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix} \) are equal if a = α, b = β, c = γ and d = δ.

2.1.1 Types of Matrices

(1) Row Matrix

A matrix with just one row is called a row matrix. For example, A = [1 5 3 -7] is a 1 × 4 row matrix.

(2) Column Matrix

A matrix with just one column is called a column matrix. For example, A =
\( \begin{bmatrix} 3 \\ 7 \\ 0 \end{bmatrix} \) is a 3 × 1 column matrix.

(3) Zero Matrix

An m × n matrix where every element is zero is called a zero matrix or a null matrix of order m × n. For example,
\( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \) is a zero matrix of order 2 × 3.

(4) Square Matrix

A matrix where the number of rows equals the number of columns, call it n, is called a square matrix of order n or an n-rowed square matrix.

The elements aij of a square matrix A = [aij] where i = j are called the diagonal elements, and the line along which these elements sit is called the principal diagonal or simply the diagonal of the matrix. Thus, the elements a11, a22, a33, …, ann are the diagonal elements of the square matrix A = [aij]n × n.

For example,
\( \begin{bmatrix} 2 & 3 & -1 \\ 4 & 7 & 9 \\ 0 & 1 & 3 \end{bmatrix} \) is a square matrix of order 3 and its diagonal elements are 2, 7, 3.

(5) Diagonal Matrix

A square matrix A = [aij]n × n is called a diagonal matrix if all its non-diagonal elements are zero. The diagonal elements themselves may or may not be zero.

For example, [4],
\( \begin{bmatrix} 3 & 0 \\ 0 & -1 \end{bmatrix} \) and \( \begin{bmatrix} 5 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \) are diagonal matrices of order 1, 2, and 3 respectively.

A diagonal matrix of order n is written as diagonal [a11, a22, a33, …, ann].

(6) Scalar Matrix

A diagonal matrix where all diagonal elements are the same is called a scalar matrix. Thus in a scalar matrix, A = [aij]n × n, we have:

aij = \( \begin{cases} 0, & \text{when } i \neq j \\ k, & \text{when } i = j \end{cases} \)

For example, [3],
\( \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \) and \( \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \) are scalar matrices of order 1, 2, and 3 respectively.

(7) Identity Matrix

A scalar matrix where each diagonal element is 1 is called an identity matrix or a unit matrix. Thus in an identity matrix, A = [aij]n × n, we have:

aij = \( \begin{cases} 0, & \text{when } i \neq j \\ 1, & \text{when } i = j \end{cases} \)

For example, [1],
\( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) are identity matrices of order 1, 2, and 3 respectively.

An identity matrix of order n is usually shown as In.

(8) Triangular Matrix

(i) Upper Triangular Matrix. A square matrix A = [aij]n × n is called an upper triangular matrix if aij = 0 for all i > j - that is, if all elements below the principal diagonal are zero.

(ii) Lower Triangular Matrix. A square matrix A = [aij]n × n is called a lower triangular matrix if aij = 0 for all i < j - that is, if all elements above the principal diagonal are zero.

(iii) Triangular Matrix. A matrix that is upper or lower triangular is called a triangular matrix. A triangular matrix A = [aij]n × n is called strictly triangular if aii = 0 for all i = 1, 2, 3, …, n.

For example,
\( \begin{bmatrix} 1 & -3 & 0 \\ 0 & 2 & 4 \\ 0 & 0 & 0 \end{bmatrix} \) and \( \begin{bmatrix} 7 & 0 & 0 \\ 4 & 0 & 0 \\ 1 & 2 & 9 \end{bmatrix} \) are upper triangular and lower triangular matrices of order 3 respectively.

Illustrative Examples

Example 1. A is a matrix of the type 3 × 5 and R is a row of A, then what is the type of R as a matrix?
Answer: Since A is a matrix of type 3 × 5, each row of A holds 5 elements. Therefore, R is a row matrix of type 1 × 5.
In simple words: When a matrix has 3 rows and 5 columns, every single row must have 5 items in it.

Exam Tip: Remember that the number of columns always tells you how many elements are in each row of a matrix.

Example 2. Let A = \( \begin{bmatrix} 3 & 7 & -1 \\ 0 & 2 & 5 \end{bmatrix} \) and B = \( \begin{bmatrix} 7 & 9 & 0 \\ 3 & -5 & 6 \end{bmatrix} \) . Find the sum of (2, 2) entries of A and B.
Answer: The (2, 2) entry of A is 2 and the (2, 2) entry of B is -5. Therefore, the sum of (2, 2) entries of A and B = 2 + (-5) = -3.
In simple words: Find the entry in row 2, column 2 of each matrix, then add them together.

Exam Tip: Always identify the correct row and column carefully - the first number in (i, j) is the row and the second is the column.

Example 3. Construct a 2 × 3 matrix A whose elements aij are given by (i) aij = 2i - 3j (ii) aij = i · j
Answer:

(i) aij = 2i - 3j, where i = 1, 2 and j = 1, 2, 3

a11 = 2(1) - 3(1) = -1, a12 = 2(1) - 3(2) = -4, a13 = 2(1) - 3(3) = -7
a21 = 2(2) - 3(1) = 1, a22 = 2(2) - 3(2) = -2, a23 = 2(2) - 3(3) = -5

Therefore, the required matrix A =
\( \begin{bmatrix} -1 & -4 & -7 \\ 1 & -2 & -5 \end{bmatrix} \)

(ii) aij = i·j, where i = 1, 2 and j = 1, 2, 3

a11 = 1·1 = 1, a12 = 1·2 = 2, a13 = 1·3 = 3
a21 = 2·1 = 2, a22 = 2·2 = 4, a23 = 2·3 = 6

Therefore, the required matrix A =
\( \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{bmatrix} \)

In simple words: Use the formula to calculate each entry by putting in the row and column numbers. Then arrange all the answers in the right positions in the matrix.

Exam Tip: Always calculate all entries systematically - first complete all entries in row 1, then move to row 2, checking your arithmetic each step.

Example 4. For what values of x and y are the following matrices equal? A = \( \begin{bmatrix} 2x + 1 & y^2 + 2 \\ 5 & y^2 - 5y \end{bmatrix} \) , B = \( \begin{bmatrix} x + 3 & 3y \\ 5 & -6 \end{bmatrix} \)
Answer: By the definition of equal matrices, we have:
2x + 1 = x + 3 ... (i)
y² + 2 = 3y ... (ii)
y² - 5y = -6 ... (iii)

From (i), 2x - x = 3 - 1
\( \implies \) x = 2

From (ii), y² - 3y + 2 = 0
\( \implies \) (y - 1)(y - 2) = 0
\( \implies \) y = 1 or y = 2

From (iii), y² - 5y + 6 = 0
\( \implies \) (y - 2)(y - 3) = 0
\( \implies \) y = 2 or y = 3

Since equations (ii) and (iii) must hold at the same time, we take the common value of y. Therefore, y = 2.

Hence, A = B if x = 2 and y = 2.
In simple words: Set each matching entry equal to each other and solve for x and y. Check which value works in all equations.

Exam Tip: When two matrices are equal, every entry in one must match the same position in the other - use this to set up equations, then solve them carefully.

Example 5. Find x, y, a and b if \( \begin{bmatrix} 3x + 4y & 6 & x - 2y \\ a + b & 2a - b & -3 \end{bmatrix} \) = \( \begin{bmatrix} 2 & 6 & 4 \\ 5 & -5 & -3 \end{bmatrix} \)
Answer: By the definition of equal matrices, we get:
3x + 4y = 2 ... (i)
x - 2y = 4 ... (ii)
a + b = 5 ... (iii)
2a - b = -5 ... (iv)

To find x, y:
Multiply (ii) by 2 and add to (i):
5x = 10
\( \implies \) x = 2

From (i): 3(2) + 4y = 2
\( \implies \) 4y = -4
\( \implies \) y = -1

To find a, b:
Add (iii) and (iv):
3a = 0
\( \implies \) a = 0

From (iii): 0 + b = 5
\( \implies \) b = 5

Hence, x = 2, y = -1, a = 0, b = 5.
In simple words: Write an equation by matching each entry, then solve the equations using addition, subtraction, or substitution.

Exam Tip: Always separate x, y variables from a, b variables into two sets of equations - solve each set independently to avoid mixing them up.

Exercise 2.1

Question 1. If A is a matrix of type p × q and R is a row of A, then what is the type of R as a matrix?
Answer: R is a row matrix of type 1 × q.
In simple words: A row stays in the same number of columns as the original matrix, but it has just one row.

Exam Tip: The type of a row is always 1 × (number of columns in the original matrix).

Question 2. If A is a column matrix with 5 rows, then what type of matrix is a row of A?
Answer: A row of a column matrix consists of just one element, so its type is 1 × 1.
In simple words: A column matrix has only one column, so any row in it contains just one item.

Exam Tip: Always remember that a row from a column matrix will be a single element arranged as 1 × 1.

Question 3. If a matrix has 12 elements, what are the possible orders it can have? What if it has 7 elements?
Answer: For a matrix with 12 elements, the possible orders are: 1 × 12, 12 × 1, 2 × 6, 6 × 2, 3 × 4, and 4 × 3. These come from all pairs of numbers that multiply to give 12.

For a matrix with 7 elements, the possible orders are: 1 × 7 and 7 × 1. This is because 7 is a prime number and can only be factored as 1 and 7.
In simple words: Find all ways to multiply two numbers to get the total number of elements. Each way gives one possible order.

Exam Tip: Remember that m × n = total elements, so find the factor pairs of the given total to list all possible matrix orders.

Question 4. Construct a 2 × 2 matrix A = [aij] whose elements are given by (i) aij = ((i + 2)²)/2 (ii) aij = (1/2)|2i - 3j|
Answer:

(i) aij = ((i + 2)²)/2, where i = 1, 2 and j = 1, 2

a11 = ((1 + 2)²)/2 = 9/2, a12 = ((1 + 2)²)/2 = 9/2
a21 = ((2 + 2)²)/2 = 8, a22 = ((2 + 2)²)/2 = 8

Therefore, A =
\( \begin{bmatrix} 9/2 & 9/2 \\ 8 & 8 \end{bmatrix} \)

(ii) aij = (1/2)|2i - 3j|, where i = 1, 2 and j = 1, 2

a11 = (1/2)|2(1) - 3(1)| = (1/2)|−1| = 1/2, a12 = (1/2)|2(1) - 3(2)| = (1/2)|−4| = 2
a21 = (1/2)|2(2) - 3(1)| = (1/2)|1| = 1/2, a22 = (1/2)|2(2) - 3(2)| = (1/2)|−2| = 1

Therefore, A =
\( \begin{bmatrix} 1/2 & 2 \\ 1/2 & 1 \end{bmatrix} \)
In simple words: Plug in each pair of i and j values into the formula, calculate carefully, and place each result in its correct spot in the matrix.

Exam Tip: Be careful with absolute value signs - find the value inside first, then take the absolute value. Check all arithmetic twice.

Question 5. Construct a 2 × 3 matrix whose elements aij are given by (i) aij = 2i - j (ii) aij = (i - j)/(i + j) (iii) aij = ((i - 2)²)/2
Answer:

(i) aij = 2i - j, where i = 1, 2 and j = 1, 2, 3

a11 = 2(1) - 1 = 1, a12 = 2(1) - 2 = 0, a13 = 2(1) - 3 = -1
a21 = 2(2) - 1 = 3, a22 = 2(2) - 2 = 2, a23 = 2(2) - 3 = 1

Therefore, A =
\( \begin{bmatrix} 1 & 0 & -1 \\ 3 & 2 & 1 \end{bmatrix} \)

(ii) aij = (i - j)/(i + j), where i = 1, 2 and j = 1, 2, 3

a11 = (1 - 1)/(1 + 1) = 0/2 = 0, a12 = (1 - 2)/(1 + 2) = -1/3, a13 = (1 - 3)/(1 + 3) = -2/4 = -1/2
a21 = (2 - 1)/(2 + 1) = 1/3, a22 = (2 - 2)/(2 + 2) = 0/4 = 0, a23 = (2 - 3)/(2 + 3) = -1/5

Therefore, A =
\( \begin{bmatrix} 0 & -1/3 & -1/2 \\ 1/3 & 0 & -1/5 \end{bmatrix} \)

(iii) aij = ((i - 2)²)/2, where i = 1, 2 and j = 1, 2, 3

a11 = ((1 - 2)²)/2 = 1/2, a12 = ((1 - 2)²)/2 = 1/2, a13 = ((1 - 2)²)/2 = 1/2
a21 = ((2 - 2)²)/2 = 0, a22 = ((2 - 2)²)/2 = 0, a23 = ((2 - 2)²)/2 = 0

Therefore, A =
\( \begin{bmatrix} 1/2 & 1/2 & 1/2 \\ 0 & 0 & 0 \end{bmatrix} \)
In simple words: For each (i, j) pair, substitute the numbers into the given formula and simplify. Arrange all answers in rows and columns.

Exam Tip: When the formula does not depend on j (like in part iii), all entries in the same row will be identical - this is a good way to check your work.

Question 6. (i) If \( \begin{bmatrix} x + y & x + 2 \\ 2x - y & 16 \end{bmatrix} \) = \( \begin{bmatrix} 8 & 5 \\ 1 & 3y + 1 \end{bmatrix} \) , then find the value of (y - x).
Answer: From matrix equality:
x + y = 8 ... (i)
x + 2 = 5 ... (ii)
2x - y = 1 ... (iii)
16 = 3y + 1 ... (iv)

From (ii): x = 3

From (i): 3 + y = 8
\( \implies \) y = 5

Check with (iii): 2(3) - 5 = 1 ✓

Therefore, y - x = 5 - 3 = 2
In simple words: Match entries in the same position to make equations. Solve for x and y, then calculate y minus x.

Exam Tip: Always verify your x and y values by checking them in all four equations before finalizing your answer.

Question 6. (ii) If \( \begin{bmatrix} 2x + y & 4x \\ 5x - 7 & 4x \end{bmatrix} \) = \( \begin{bmatrix} 7 & 7y - 13 \\ y & x + 6 \end{bmatrix} \) , then find the values of x and y.
Answer: From matrix equality:
2x + y = 7 ... (i)
4x = 7y - 13 ... (ii)
5x - 7 = y ... (iii)
4x = x + 6 ... (iv)

From (iv): 4x - x = 6
\( \implies \) 3x = 6
\( \implies \) x = 2

From (iii): y = 5(2) - 7 = 3

Verify with (i): 2(2) + 3 = 7 ✓
Verify with (ii): 4(2) = 7(3) - 13
\( \implies \) 8 = 8 ✓

Therefore, x = 2 and y = 3.
In simple words: Use the simplest equation first to find one variable, then use another equation to find the second variable. Check all equations to make sure.

Exam Tip: Look for the simplest equations first (ones with fewer terms) - solving them will be easier and give you quick values to work with.

Question 7. If \( \begin{bmatrix} a + b & 2 \\ 5 & ab \end{bmatrix} \) = \( \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix} \) , find the values of a and b.
Answer: From matrix equality:
a + b = 6 ... (i)
ab = 8 ... (ii)

From (i): b = 6 - a

Substitute into (ii): a(6 - a) = 8
\( \implies \) 6a - a² = 8
\( \implies \) a² - 6a + 8 = 0
\( \implies \) (a - 2)(a - 4) = 0
\( \implies \) a = 2 or a = 4

If a = 2, then b = 6 - 2 = 4
If a = 4, then b = 6 - 4 = 2

Therefore, a = 2, b = 4 or a = 4, b = 2.
In simple words: Match entries to get two equations. Use substitution to turn it into one equation with one variable, then solve.

Exam Tip: When you have sum and product of two numbers, think about factoring a quadratic - it usually leads to the answer quickly.

Question 8. If \( \begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} \) = \( \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \) , find x, y, z and w.
Answer: From matrix equality:
x - y = -1 ... (i)
2x + z = 5 ... (ii)
2x - y = 0 ... (iii)
3z + w = 13 ... (iv)

From (i) and (iii):
Subtracting (i) from (iii): x = 1

From (i): 1 - y = -1
\( \implies \) y = 2

From (ii): 2(1) + z = 5
\( \implies \) z = 3

From (iv): 3(3) + w = 13
\( \implies \) w = 4

Therefore, x = 1, y = 2, z = 3, w = 4.
In simple words: Write all four equations from matching entries, then solve step-by-step starting with the simplest equations.

Exam Tip: When you have multiple equations, subtracting similar equations from each other often eliminates variables quickly and gives you an answer faster.

Question 9. Find x, y, a and b if \( \begin{bmatrix} 2x - 3y & a - b & 3 \\ 1 & x + 4y & 3a + 4b \end{bmatrix} \) = \( \begin{bmatrix} 1 & -2 & 3 \\ 1 & 6 & 29 \end{bmatrix} \)
Answer: From matrix equality:
2x - 3y = 1 ... (i)
a - b = -2 ... (ii)
x + 4y = 6 ... (iii)
3a + 4b = 29 ... (iv)

To find x, y:
From (i): 2x - 3y = 1
From (iii): x + 4y = 6
\( \implies \) x = 6 - 4y

Substitute into (i): 2(6 - 4y) - 3y = 1
\( \implies \) 12 - 8y - 3y = 1
\( \implies \) -11y = -11
\( \implies \) y = 1

From (iii): x = 6 - 4(1) = 2

To find a, b:
From (ii): a = b - 2

Substitute into (iv): 3(b - 2) + 4b = 29
\( \implies \) 3b - 6 + 4b = 29
\( \implies \) 7b = 35
\( \implies \) b = 5

From (ii): a = 5 - 2 = 3

Therefore, x = 2, y = 1, a = 3, b = 5.
In simple words: Separate the variables into two groups - x, y in one group and a, b in another. Solve each group separately.

Exam Tip: Always solve independent groups of variables separately - it reduces confusion and makes checking your work easier.

Question 10. If \( \begin{bmatrix} x + 3 & z + 4 & 2y - 7 \\ 4x + 6 & a - 1 & 0 \\ b - 3 & 3b & z + 2c \end{bmatrix} \) = \( \begin{bmatrix} 0 & 6 & 3y - 2 \\ 2x & -3 & 2c + 2 \\ 2b + 4 & -21 & 0 \end{bmatrix} \) , find the values of a, b, c, x, y and z.
Answer: From matrix equality, matching all corresponding entries:

x + 3 = 0
\( \implies \) x = -3

z + 4 = 6
\( \implies \) z = 2

2y - 7 = 3y - 2
\( \implies \) -y = 5
\( \implies \) y = -5

4x + 6 = 2x
\( \implies \) 2x = -6
\( \implies \) x = -3 (confirms above)

a - 1 = -3
\( \implies \) a = -2

b - 3 = 2b + 4
\( \implies \) -b = 7
\( \implies \) b = -7

3b = -21
\( \implies \) b = -7 (confirms above)

z + 2c = 0
\( \implies \) 2 + 2c = 0
\( \implies \) c = -1

Therefore, a = -2, b = -7, c = -1, x = -3, y = -5, z = 2.
In simple words: Write out an equation for each matching entry from both matrices, then solve each equation. When the same variable appears in multiple equations, use those to verify your answer.

Exam Tip: When a variable appears in more than one equation, solve it from all equations - if you get different answers, you made a calculation error and need to recheck.

2.2 Operations on Matrices

2.2.1 Addition of Matrices

If A and B are two matrices of the same order, we say that these are compatible (or conformable

Thus, if A = [aij]m × n and B = [bij]m × n, then A + B = [aij + bij]m × n, where 1 ≤ i ≤ m, 1 ≤ j ≤ n.

For example, if A =
\( \begin{bmatrix} 2 & 3 & -1 \\ 5 & 0 & 7 \end{bmatrix} \) and B =
\( \begin{bmatrix} 4 & 9 & -11 \\ 3 & 1 & -2 \end{bmatrix} \)

then A + B =
\( \begin{bmatrix} 2 + 4 & 3 + 9 & -1 + (-11) \\ 5 + 3 & 0 + 1 & 7 + (-2) \end{bmatrix} \) =
\( \begin{bmatrix} 6 & 12 & -12 \\ 8 & 1 & 5 \end{bmatrix} \)

Note: If A and B are not of the same order, then A + B is not defined, and we say that A and B are not compatible for the sum A + B.

2.2.2 Subtraction of Matrices

If A and B are two matrices of the same order, we say that these are compatible for subtraction, and their difference A - B is the matrix obtained by subtracting the elements of B from the matching elements of A.

Thus, if A = [aij]m × n and B = [bij]m × n, then A - B = [aij - bij]m × n, where 1 ≤ i ≤ m, 1 ≤ j ≤ n.

For example, if A =
\( \begin{bmatrix} 4 & -7 & -2 \\ 5 & 3 & 0 \end{bmatrix} \) and B =
\( \begin{bmatrix} 2 & 4 & -5 \\ -9 & 6 & -3 \end{bmatrix} \)

then A - B =
\( \begin{bmatrix} 4 - 2 & -7 - 4 & -2 - (-5) \\ 5 - (-9) & 3 - 6 & 0 - (-3) \end{bmatrix} \) =
\( \begin{bmatrix} 2 & -11 & 3 \\ 14 & -3 & 3 \end{bmatrix} \)

Properties of Matrix Addition

(1) Addition of matrices is commutative

That is, if A and B are matrices of the same order, then A + B = B + A.

Proof: Let A = [aij]m × n and B = [bij]m × n be two matrices of the same order m × n, then:

A + B = [aij]m × n + [bij]m × n = [aij + bij]m × n (by definition of sum)
= [bij + aij]m × n (since addition of numbers is commutative)
= [bij]m × n + [aij]m × n = B + A.

(2) Addition of matrices is associative

That is, if A, B and C are matrices of the same order, then (A + B) + C = A + (B + C).

Proof: Let A = [aij]m × n, B = [bij]m × n and C = [cij]m × n be three matrices of the same order, then:

(A + B) + C = ([aij]m × n + [bij]m × n) + [cij]m × n
= [aij + bij]m × n + [cij]m × n
= [(aij + bij) + cij]m × n
= [aij + (bij + cij)]m × n (since addition of numbers is associative)
= [aij]m × n + [bij + cij]m × n
= [aij]m × n + ([bij]m × n + [cij]m × n)
= A + (B + C).

(3) Existence of Additive Identity

For any matrix A, there is a null matrix O of the same order such that A + O = A = O + A.

Proof: Let A = [aij]m × n be any matrix and O be the null matrix of the same order m × n, then:

O = [0ij]m × n where 0ij = 0 for all i, j.

Now A + O = [aij]m × n + [0ij]m × n = [aij + 0ij]m × n
= [aij + 0]m × n = [aij]m × n (since a + 0 = a for all numbers a)
= A

Also, A + O = O + A (since addition of matrices is commutative)
∴ A + O = A = O + A.

(4) Existence of Additive Inverse

For any matrix A = [aij]m × n, there is a matrix - A = [- aij]m × n such that A + (- A) = O = (- A) + A.

Proof: A + (- A) = [aij]m × n + [- aij]m × n
= [aij + (- aij)]m × n = [0]m × n = [0ij]m × n = Om × n

Also, (- A) + A = A + (- A) (since addition of matrices is commutative)
∴ A + (- A) = O = (- A) + A.

The matrix - A = [- aij]m × n is called the additive inverse of the matrix A = [aij]m × n.

For example, if A =
\( \begin{bmatrix} 2 & -3 & 0 \\ 5 & -7 & 6 \end{bmatrix} \), then - A =
\( \begin{bmatrix} -2 & 3 & 0 \\ -5 & 7 & -6 \end{bmatrix} \)

2.2.3 Multiplication of a matrix by a scalar (number)

If k is any scalar (number) and A is any matrix, then the matrix obtained by multiplying each element of A by k is called the scalar multiple of A by k and is shown as k A.

Thus, if A = [aij]m × n, then k A = [k aij]m × n.

For example, if A =
\( \begin{bmatrix} 2 & 5 & 0 \\ -3 & 7 & 9 \end{bmatrix} \)

then 3A =
\( \begin{bmatrix} 3 \times 2 & 3 \times 5 & 3 \times 0 \\ 3 \times (-3) & 3 \times 7 & 3 \times 9 \end{bmatrix} \) =
\( \begin{bmatrix} 6 & 15 & 0 \\ -9 & 21 & 27 \end{bmatrix} \)

Properties of Scalar Multiplication

If A, B are two matrices of the same order and k, l are any scalars (numbers), then:

(i) (k + l) A = k A + l A
(ii) k (A + B) = k A + k B
(iii) (k l) A = k (l A) = l (k A)
(iv) 1A = A
(v) (-1) A = - A