Matrix operations mainly involve three algebraic operations: addition, subtraction, and multiplication of matrices. A matrix is a rectangular array of numbers or expressions arranged in rows and columns. It has many important applications in mathematics.

Before understanding the operations on matrices, it is important to be familiar with the elementary operation of a matrix. This article covers the operations on matrices along with their properties and solved examples.

Learn More:

Introduction to Matrices

Types of Matrices

Adjoint and Inverse of a Matrix

Rank of a Matrix and Special Matrices

Operations on Matrices

The basic operations on a matrix are addition, subtraction and multiplication. To add or subtract matrices, they must be of the same order. For multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix.

Addition of Matrices

Subtraction of Matrices

Scalar Multiplication of Matrices

Multiplication of Matrices

Addition of Matrices

The sum of two matrices A[aij]mxn and B[bij]mxn of the same order is a matrix, where each element is the sum of the corresponding elements of A and B. That is, A + B = [aij + bij]mxn.

The sum of matrices A and B of order 2 x 2 is given by:

(\begin{array}{l}\begin{bmatrix} a_1 & b_1 \ c_1 & d_1 \end{bmatrix} + \begin{bmatrix} a_2 & b_2 \ c_2 & d_2 \end{bmatrix} = \begin{bmatrix} a_1+a_2 & b_1+b_2 \ c_1+c_2 & d_1+d_2 \end{bmatrix}\end{array} )

Properties of Matrix Addition: If a, B and C are matrices of same order, then

A + B = B + A

Commutative Law: A + B = B + A

Associative Law: A + (B + C) = (A + B) + C

Identity of the Matrix: A + O = O + A = A, where O is the zero matrix which is the additive identity of the matrix.

Additive Inverse: For a matrix A, its additive inverse is obtained by changing the sign of every element of A, such that A + (-A) = 0 = (-A) + A.

$$\left. \begin{array}{l}\begin{matrix} A+B=A+C \ B+A=C+A \end{matrix} \right}\Rightarrow B=C\end{array}$$

(f) tr(A ± B) = tr(A) ± tr(B)

If A + B = 0 = B + A, then B is called the additive inverse of A and A is called the additive inverse of B.

Subtraction of Matrices

If A and B are two matrices of the same order, then we define $$A-B = A + (-B).$$

The difference between matrices A and B of order 2 x 2 is given by:

(\begin{array}{l}\begin{bmatrix} a_1 &b_1 \ c_1 &d_1 \end{bmatrix} - \begin{bmatrix} a_2&b_2 \ c_2& d_2 \end{bmatrix} = \begin{bmatrix} a_1-a_2 &\ b_1-b_2 \ c_1-c_2 &\ d_1-d_2 \end{bmatrix}\end{array})

We can subtract the matrices by subtracting each element of one matrix from the corresponding element of the second matrix, i.e. A - B = [a_{ij} - b_{ij}]_{mxn}

Scalar Multiplication of Matrices

The scalar multiplication of a matrix $A = [aij]m×n$ by a scalar $k$ is denoted by $kA$ and is obtained by multiplying each element of $A$ by $k$.

Then $$kA_{m \times n} = A_{m \times n}k = \left[ ka_{i \times j} \right]$$

Properties of Scalar Multiplication: If A, B are matrices of the same order and λ and μ are any two scalars, then:

(a) λ(A + B) = λA + λB

$(b)\quad(\lambda + \mu)A = \lambda A + \mu A$

(c) λ(μA) = (λμA) = μ(λA)

(d) $(-\lambda A) = -(\lambda A) = \lambda(-A)$

$k \cdot \text{tr}(A) = \text{tr}(kA)$

Multiplication of Matrices

If A and B are any two matrices, then their product AB can only be defined when the number of columns in A is equal to the number of rows in B.

If $$A=\begin{bmatrix} a_{ij} \end{bmatrix}{m \times n} ;;;\text{and};;; B=\begin{bmatrix} b{ij} \end{bmatrix}{n \times p} ;;;\text{then their product}; AB=C=\begin{bmatrix} c{ij} \end{bmatrix}{m \times p}$$ will be a matrix of order $m \times p$ where $$\begin{bmatrix} (AB){ij} \end{bmatrix} = \begin{bmatrix} C_{ij} \end{bmatrix} = \sum_{r=1}^{n} a_{ir} b_{rj}$$

Properties of Matrix Multiplication

In general, matrix multiplication is not commutative, i.e. (AB \ne BA).

Matrix multiplication is associative, meaning that $$(AB)C = A(BC)$$.

Matrix multiplication is distributive over matrix addition, i.e. A.(B + C) = A.B + A.C and (A + B)C = AC + BC.

If A is an m × n matrix, then $${{I}{m}}A=A=A{{I}{n}}.$$

The product of two matrices can be a null matrix while neither of them is null, i.e. if AB = 0, it is not necessary that either A = 0 or B = 0.

If (A_{m\times n}) is an m × n matrix and (O_{n\times p}) is a null matrix, then (A_{m\times n} \cdot O_{n\times p} = O_{m\times p}), i.e. the product of the matrix with a null matrix is always a null matrix.

If AB = 0, it does not necessarily mean that A = 0 or B = 0, as the product of two non-zero matrices may still be a zero matrix.

If AB = AC, but B ≠ C, then the Cancellation Law is not applicable.

(i) tr(BA) = tr(AB)

There exists a multiplicative identity for every square matrix, such that $A I = I A = A$.

Matrix Operations Examples

Illustration 1: (\begin{array}{l} \text{If}\ A=\left[ \begin{matrix} 2 & 1 & 3 \ 3 & -2 & 1 \ -1 & 0 & 1 \ \end{matrix} \right]\ \text{and}\ B=\left[ \begin{matrix} 1 \ 2 \ 4 \ \end{matrix},,,,\begin{matrix} -2 \ 1 \ -2 \ \end{matrix} \right]\end{array} )

Find AB and BA if possible.

Given:

This is a statement

Solution:

This is a statement

The product of matrices A and B is of order 3 x 2, since A is a 3 x 3 matrix and B is a 3 x 2 matrix, and they are conformable for matrix multiplication.

$\begin{array}{l}(A_{1,1}) \cdot (B_{1,1})\=\left[ 2,1,3 \right]\left[ \begin{matrix} 1 \ 2 \ 4 \ \end{matrix} \right]\=2\times 1+1\times 2+3\times 4\=16\end{array}$

(\begin{array}{l}=\left[ \begin{matrix} 2 & 1 & 3 \end{matrix} \right]\left[ \begin{matrix} -2 \ 1 \ -3 \ \end{matrix} \right] \ =2\times \left( -2 \right)+1\times 1+3\times \left( -3 \right) \ =-12\end{array} )

(\begin{array}{l} (Second\ row\ of\ A)\times (First\ column\ of\ B) \=\left[ 3,-2,,1 \right]\left[ \begin{matrix} 1 \ 2 \ 4 \ \end{matrix} \right] \=3\times 1+\left( -2 \right)\times 2+1\times 4 \=3 \end{array})

Similarly $$\begin{array}{l}{\left( AB \right){22}}=-10,{\left( AB \right){31}}=3 ;and ;{\left( AB \right)_{32}}=0\end{array}$$

(\therefore AB = \left[ \begin{matrix} 16 & -12 \ 3 & -10 \ 3 & 0 \ \end{matrix} \right])

BA is not possible since the number of columns of B does not equal the number of rows of A.

Illustration 2: Find the value of x and y if (\begin{array}{l}2\left[ \begin{matrix} 1 & 3 \ 0 & x \ \end{matrix} \right]+\left[ \begin{matrix} y & 0 \ 1 & 2 \ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \ 1 & 8 \ \end{matrix} \right]\end{array} )

Solution: x = 6 and y = 3

Given:

This is a heading

Solution:

This is a heading

Using the method of multiplication and addition of matrices, then equating the corresponding elements of L.H.S. and R.H.S., we can easily get the required values of x and y.

We have, (\begin{array}{l}\left[ \begin{matrix} 2+y & 6 \ 0+1 & 2x+2 \ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \ 1 & 8 \ \end{matrix} \right]\\Rightarrow 2+y=5,\quad 6=6,\quad 0+1=1,\quad 2x+2=8\\Rightarrow y=3,\quad x=3\end{array} )

Equating the corresponding elements, a11 and a22, yields

(\begin{array}{l}y=3;,,,x=3\end{array})

Hence, x = 3 and y = 3.

Illustration 3: Find the value of a, b, c and d, if (\begin{array}{l}\left[ \begin{matrix} a-b & 2a+c \ 2a-b & 3c+d \ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5 \ 0 & 13 \ \end{matrix} \right]\end{array} )?

Given:

This is bold

Solution:

This is bold

As the two matrices are equal, their corresponding elements are equal. Hence, by equating the corresponding elements of the given matrices, we can determine the values of a, b, c, and d.

(\begin{array}{l}\left[ \begin{matrix} a-b & 2a+c \ 2a-b & 3c+d \ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5 \ 0 & 13 \ \end{matrix} \right]\end{array} )

It’s time to go!

It’s time to go!

a - b = -1 ... (i)

2a + c = 5 \\ (ii)

2a + c = 5 \\ (ii)

2a - b = 0 … (iii)

2a - b = 0 … (iii)

3c + d = 13 \\ (iv)

Subtracting equation (iii) from (i), we have a = 1;

Putting the value of $a$ in equation (i), we get $b = 2$

Putting the value of $a$ in equation (ii), we have $\begin{array}{l}2+c=5 \Rightarrow c=3;\end{array}$

Putting the value of $c$ in equation (iv), we find $\begin{array}{l}9+x=13 \Rightarrow x=4\end{array}$

Therefore, a=1, b=2, c=3, d=4.

Illustration 4: Find $x$ and $y$, if $$\begin{array}{l}2x+3y=\left[ \begin{matrix} 2 & 3 \ 4 & 0 \ \end{matrix} \right] and 3x+2y=\left[ \begin{matrix} 2 & -2 \ -1 & 5 \ \end{matrix} \right].\end{array}$$

Given:

This is a Heading

Solution:

This is a Heading

By solving the given equations simultaneously, we will be able to find the values of x and y.

We have $$2x+3y=\left[ \begin{matrix} 2 & 3 \ 4 & 0 \ \end{matrix} \right]….(i)$$

(\begin{array}{l}3x+2y=\left[ \begin{matrix} 2 & -2 \ -1 & 5 \ \end{matrix} \right] \Rightarrow \left[ \begin{matrix} x \ y \ \end{matrix} \right] = \frac{1}{11}\left[ \begin{matrix} 23 & 2 \ 7 & -10 \ \end{matrix} \right]\end{array} )

Multiplying (i) by 3 and (ii) by 2, we get (\begin{array}{l}6x+9y=\left[ \begin{matrix} 18 & 27 \ 24 & 0 \ \end{matrix} \right]….(iii)\end{array} )

(\begin{array}{l}6x+4y=\left[ \begin{matrix} 4 & -4 \ -2 & 10 \ \end{matrix} \right] \Rightarrow \left[ \begin{matrix} x \ y \ \end{matrix} \right] = \frac{1}{20}\left[ \begin{matrix} 10 & 4 \ 4 & -6 \ \end{matrix} \right]\left[ \begin{matrix} 4 & -4 \ -2 & 10 \ \end{matrix} \right] \end{array})

Subtracting (iv) from (iii), we get (\begin{array}{l}5y=\left[ \begin{matrix} 6-4-4 & 9+4-4 \ 12+2-4 & 0-10-4 \ \end{matrix} \right]=\left[ \begin{matrix} 2 & 13 \ 14 & -10 \ \end{matrix} \right]\end{array} )

$\Rightarrow y=\left[ \begin{matrix} \frac{2}{5} & \frac{13}{5} \ \frac{14}{5} & -2 \ \end{matrix} \right]$

Putting the value of y in (iii), we get (\begin{array}{l}2x+3\left[ \begin{matrix} \frac{2}{5} & \frac{13}{5} \ \frac{14}{5} & -2 \ \end{matrix} \right]=\left[ \begin{matrix} 2 & 3 \ 4 & y \ \end{matrix} \right]\end{array} )

(\begin{array}{l} \Rightarrow 2x=\left[ \begin{matrix} 2 & 3 \ 4 & 0 \ \end{matrix} \right] - \left[ \begin{matrix} \frac{6}{5} & \frac{39}{5} \ \frac{42}{5} & -6 \ \end{matrix} \right]\ =\left[ \begin{matrix} 2-\frac{6}{5} & 3-\frac{39}{5} \ 4-\frac{42}{5} & 0+6 \ \end{matrix} \right]\ =\left[ \begin{matrix} \frac{4}{5} & -\frac{24}{5} \ -\frac{22}{5} & 6 \ \end{matrix} \right]\ \Rightarrow x=\left[ \begin{matrix} \frac{2}{5} & -\frac{12}{5} \ -\frac{11}{5} & 3 \ \end{matrix} \right] \end{array})

Hence $$x=\begin{bmatrix} \frac{2}{5} & -\frac{12}{5} \ -\frac{11}{5} & 3 \end{bmatrix} ;;\text{and};; y=\begin{bmatrix} \frac{2}{5} & \frac{13}{5} \ \frac{14}{5} & -2 \end{bmatrix}$$

Frequently Asked Questions

Yes, matrix addition is commutative.

Yes, Matrix addition is commutative.

No, matrix multiplication is not commutative.

No, Matrix multiplication is not commutative.

Yes, matrix addition is associative.

Yes, matrix addition is associative.

What is the product of a matrix with a null matrix?

The product of a matrix with a null matrix is a null matrix.

The product of a matrix and a null matrix is a null matrix.