In mathematics, we might have come across performing different arithmetic operations on numbers, functions, polynomials, etc. Similarly, we can perform arithmetic operations such as addition, subtraction and multiplication on matrices. However, division of matrices is not possible. In this article, you will learn what is addition of matrices and how to add the given matrices, properties and solved examples.

Get the basics of Matrices here which will help in understanding the algebra of matrices.

If $A = [a_{ij}]$ and $B = [b_{ij}]$ are two matrices of the same order, say m Ã— n, then the sum of the two matrices A and B is defined as a matrix $C = [c_{ij}]_{m\times n}$, where $c_{ij} = a_{ij} + b_{ij}$, for all possible values of i and j.

i.e. C = A + B

Suppose $A = \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ a_{31} & a_{32} \end{bmatrix}$ and $B = \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22}\\ b_{31} & b_{32} \end{bmatrix}$ are two matrices of order 3 x 2 such that the sum of these two matrices is given by:

C = A + B

Therefore,

$C=\begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ a_{31} & a_{32} \end{bmatrix}+ \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22}\\ b_{31} & b_{32} \end{bmatrix}=\begin{bmatrix} a_{11}+b_{11} & a_{12}+b_{12}\\ a_{21}+b_{21} & a_{22}+b_{22}\\ a_{31}+b_{31} & a_{32}+b_{32} \end{bmatrix}$

That means the sum or addition of two matrices is a matrix obtained by adding the corresponding elements of the given two matrices. Also, it is essential to note that the two matrices have to be of the same order.

### How To Do Matrix Addition?

To add matrices, it is necessary that they have the same dimensions, i.e. the order of the matrices must be the same. Letâ€™s have a look at the example given below to learn how to add two matrices.

Example: If $X =\begin{bmatrix} 1 &-1 & 2\\ 0 & 3 & 4 \end{bmatrix}$ and $Y =\begin{bmatrix} 2 &-1 & 5\\ 7 & 1 & 4 \end{bmatrix}$, then find X + Y.

Solution:

$X =\begin{bmatrix} 1 &-1 & 2\\ 0 & 3 & 4 \end{bmatrix}$

and

$Y =\begin{bmatrix} 2 &-1 & 5\\ 7 & 1 & 4 \end{bmatrix}$

Here, the order of the matrix X is 2 x 3 and the order of Y is 2 x 3 are the same. So, we can add the given two matrices by adding the corresponding elements.

\begin{align*} X + Y &=\begin{bmatrix} 1 &-1 & 2\\ 0 & 3 & 4 \end{bmatrix} +\begin{bmatrix} 2 &-1 & 5\\ 7 & 1 & 4 \end{bmatrix}\\&=\begin{bmatrix} 1+2 &-1+(-1) & 2+5\\ 0+7 & 3+1 & 4+4 \end{bmatrix} \\&=\begin{bmatrix} 3 &-2 & 7\\ 7 & 4 & 8 \end{bmatrix} \end{align*}

Similarly, we can find the sum of two matrices with the same order.

Suppose $A=\begin{bmatrix} 2 &-1\\ 5 & 6\end{bmatrix}$ and $B=\begin{bmatrix} 3 &0\\ 1 &-1\\4 & 5\end{bmatrix}$, then the addition of A and B is not possible since the order of matrix A is 2 x 2 and the order of B is 2 x 3, i.e. the order of these matrices is not equal.

## Properties of Addition of Matrices

Below are the properties of addition of matrices.

Commutative Law: If A = [aij], B = [bij] are matrices of the same order, say m Ã— n, then:

A + B = B + A

i.e. A + B = [aij] + [bij]

= [aij + bij]

= [bij + aij] (addition of numbers is commutative)

= ([bij] + [aij])

= B + A

Therefore, the addition of matrices is commutative.

Associative Law: For any three matrices A = [aij], B = [bij], C = [cij] of the same order, say m Ã— n, then:

(A + B) + C = A + (B + C)

This can be shown as:

(A + B) + C = ([aij] + [bij]) + [cij]

= [aij + bij] + [cij]

= [(aij + bij) + cij]

= [aij + (bij + cij)] (addition of numbers is associative)

= [aij] + [(bij + cij)]

= [aij] + ([bij] + [cij])

= A + (B + C)

Existence of additive identity: Let A = [aij] be an m Ã— n matrix and O be an m Ã— n zero matrix, then:

A + O = O + A = A

The existence of additive inverse: Let A = [aij] be any matrix of the order m Ã— n, then we have another matrix as â€“A = [â€“aij]m Ã— n such that A + (â€“A) = (â€“A) + A = O.

Thus, â€“A is the additive inverse of A or negative of A. The negative of a matrix is denoted by â€“A and it can be defined as â€“A = (â€“1) A.

Question 1: If $A=\begin{bmatrix} 2 &3 \\ 4 & 5\\1 &6\end{bmatrix}$ and $B=\begin{bmatrix} 3 &-1 \\ -1 & 0\\3 &2\end{bmatrix}$, then find A + B and B + A.

Solution:

Given,

$A=\begin{bmatrix} 2 &3 \\ 4 & 5\\1 &6\end{bmatrix}$

and

$B=\begin{bmatrix} 3 &-1 \\ -1 & 0\\3 &2\end{bmatrix}$

Now,

\begin{align*}A+B &=\begin{bmatrix} 2 &3 \\ 4 & 5\\1 &6\end{bmatrix}+\begin{bmatrix} 3 &-1 \\ -1 & 0\\3 &2\end{bmatrix}\\ & =\begin{bmatrix} 2+3 &3+(-1) \\ 4+(-1) & 5+0\\1+3 &6+2\end{bmatrix}\\&=\begin{bmatrix} 5 &2 \\ 3 & 5\\4 &8\end{bmatrix}\end{align*}

Let us calculate B + A as:

\begin{align*}B + A &=\begin{bmatrix} 3 &-1 \\ -1 & 0\\3 &2\end{bmatrix}+\begin{bmatrix} 2 &3 \\ 4 & 5\\1 &6\end{bmatrix}\\ & =\begin{bmatrix} 3+2 &-1+3 \\ -1+4 & 0+5\\3+1 &2+6\end{bmatrix}\\&=\begin{bmatrix} 5 &2 \\ 3 & 5\\4 &8\end{bmatrix}\end{align*}

Therefore, A + B = B + A.

Question 2: If $\begin{bmatrix} 2 &6 \\ 0 & 2x \end{bmatrix}+\begin{bmatrix} y &0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 &6 \\ 1 & 8 \end{bmatrix}$, then find x and y.

Solution:

Given,

$\begin{bmatrix} 2 &6 \\ 0 & 2x \end{bmatrix}+\begin{bmatrix} y &0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 &6 \\ 1 & 8 \end{bmatrix}$

By performing the addition of matrices, we get;

$\begin{bmatrix} 2+y &6+0 \\ 0+1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 &6 \\ 1 & 8 \end{bmatrix}$

i.e. $\begin{bmatrix} 2+y &6 \\ 1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 &6 \\ 1 & 8 \end{bmatrix}$

Now, by equating the corresponding elements,

2x + 2 = 8

â‡’ 2x = 6

â‡’ x = 3

Also, 2 + y = 5

â‡’ y = 3

Therefore, x = 3 and y = 3.

Question 3: Find the sum of $A = \begin{bmatrix} 3 &-1 &2\\ 4 &2 &5\\ 2 &0 &3 \end{bmatrix}$ and $B = \begin{bmatrix} -12 &7 &6\\ 8 &0 &5\\ 3 &2 &-4 \end{bmatrix}$.

Solution:

Given,

A = \begin{bmatrix} 3 &-1 &2\\ 4 &2 &5\\ 2 &0 &3 \end{bmatrix}\)

and

$B = \begin{bmatrix} -12 &7 &6\\ 8 &0 &5\\ 3 &2 &-4 \end{bmatrix}$

Sum of A and B is:

\begin{align*}A + B &= \begin{bmatrix} 3 &-1 &2\\ 4 &2 &5\\ 2 &0 &3 \end{bmatrix}+ \begin{bmatrix} -12 &7 &6\\ 8 &0 &5\\ 3 &2 &-4 \end{bmatrix}\\&=\begin{bmatrix} 3+(-12) &-1+7 &2+6\\ 4+8 &2+0 &5+5\\ 2+3 &0+2 &3+(-4) \end{bmatrix}\\&=\begin{bmatrix} -9 &6 &8\\ 12 &2 &10\\ 5 &2 &-1 \end{bmatrix}\end{align*}