Just like operations on numbers, matrix addition and multiplication follows some properties as follows:

• Commutative Property of Matrix Addition
• Associative Property of Matrix Addition
• Associative Property of Multiplication
• Identity Property of Matrix
• Inverse Property of Matrix
• Distributive Property of Matrix Addition
• Distributive Property of Scalar Addition

Matrix operations can be performed by following the above properties.

## Commutative Law of Matrix Addition

Communicative law of addition for matrices is stated as follows:

If matrices A and B are of same order then, A + B = B + A.

Proof:

Let A = $\begin{bmatrix} 2 &-3 &4 \\ -1 &6 &-5 \\ 1 &3 & 5 \end{bmatrix}$ and B = $\begin{bmatrix} 1 &7 &9 \\ -3 &8 &0 \\ 2 &1 & 1 \end{bmatrix}$

Now, A + B = $\begin{bmatrix} 2 &-3 &4 \\ -1 &6 &-5 \\ 1 &3 & 5 \end{bmatrix}$ + $\begin{bmatrix} 1 &7 &9 \\ -3 &8 &0 \\ 2 &1 & 1 \end{bmatrix}$

In the addition of matrices, we can add only the corresponding elements.
Hence,
A + B = $\begin{bmatrix} 3 &4 &13 \\ -4 &14 &-5 \\ 3 &4 & 6 \end{bmatrix}$

Again,
B + A = $\begin{bmatrix} 1 &7 &9 \\ -3 &8 &0 \\ 2 &1 & 1 \end{bmatrix}$ + $\begin{bmatrix} 2 &-3 &4 \\ -1 &6 &-5 \\ 1 &3 & 5 \end{bmatrix}$

= $\begin{bmatrix} 3 &4 &13 \\ -4 &14 &-5 \\ 3 &4 & 6 \end{bmatrix}$

$\Rightarrow$ A + B = B + A i.e.addition of matrices is commutative.

## Associative Law of Matrix Addition

Associative law of addition for matrices is stated as follows:

Any three matrices of same order A, B and C, we have (A + B) + C = A + ( B + C).

Proof:

Let A, B and C are the same order (m x n) matrices defined as
A = $\begin{bmatrix} 2 &-8 \\ 4 &-5 \end{bmatrix}$, B = $\begin{bmatrix} 3 &-4 \\ 6 & 1 \end{bmatrix}$, C = $\begin{bmatrix} 1 &9 \\ 5 &7 \end{bmatrix}$

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

= $\begin{bmatrix} 5 &-12 \\ 10 &-4 \end{bmatrix}$

Again,
(A + B) + C = $\begin{bmatrix} 5 &-12 \\ 10 &-4 \end{bmatrix}$ + $\begin{bmatrix} 1 &9 \\ 5 &7 \end{bmatrix}$

= $\begin{bmatrix} 6 &-3 \\ 15 &3 \end{bmatrix}$ ...................................(i)

(B + C) = $\begin{bmatrix} 3 &-4 \\ 6 & 1 \end{bmatrix}$ + $\begin{bmatrix} 1 &9 \\ 5 &7 \end{bmatrix}$

= $\begin{bmatrix} 4 &5 \\ 11 &8 \end{bmatrix}$

A + (B + C) = $\begin{bmatrix} 2 &-8 \\ 4 &-5 \end{bmatrix}$ + $\begin{bmatrix} 4 &5 \\ 11 &8 \end{bmatrix}$

= $\begin{bmatrix} 6 &-3 \\ 15 &3 \end{bmatrix}$ ......................................(ii)

From (i) and (ii), we get (A + B) + C = A + ( B + C).

Additive Identity for matrices is stated as follows:

If Amxn = [aij] is a matrix and 0mxn is a null matrix then, A + 0 = A = 0 + A.

Proof:
Let A2x3 = $\begin{bmatrix} 3 &5 &-2 \\ 1 &7 &-6 \end{bmatrix}$ and 02x3 = $\begin{bmatrix} 0 &0 &0\\ 0&0 &0 \end{bmatrix}$

Now, A + 0 = $\begin{bmatrix} 3 &5 &-2 \\ 1 &7 &-6 \end{bmatrix}$ + $\begin{bmatrix} 0 &0 &0\\ 0&0 &0 \end{bmatrix}$

= $\begin{bmatrix} 3 &5 &-2 \\ 1 &7 &-6 \end{bmatrix}$ = A ......................................(a)

Again 0 + A = $\begin{bmatrix} 0 &0 &0 \\ 0 &0 &0 \end{bmatrix}$ + $\begin{bmatrix} 3 &5 &-2 \\ 1 &7 &-6 \end{bmatrix}$

= $\begin{bmatrix} 3 &5 &-2 \\ 1 &7 &-6 \end{bmatrix}$ = A .......................................(b)

It is clear from (a) and (b) that A + 0 = A = 0 + A

In matrices, 0(null matrix) is the additive identity. So, the additive inverse of a matrix A is the matrix's negative or in other words the additive inverse of a matrix A is the matrix to which when A is added we get Null matrix(0). The existence of additive Inverse is stated as follows:

A + (-A) = 0 = (-A) + A

Proof:

If A = $\begin{bmatrix} a &b \\ -c &d \\ e &-f \end{bmatrix}$ is the given matrix,

so -A = - $\begin{bmatrix} a &b \\ -c &d \\ e &-f \end{bmatrix}$

= $\begin{bmatrix} -a &-b \\ c &-d \\ -e &f \end{bmatrix}$

Now A + (-A) = $\begin{bmatrix} a &b \\ -c &d \\ e &-f \end{bmatrix}$ + $\begin{bmatrix} -a &-b \\ c &-d \\ -e &f \end{bmatrix}$

= $\begin{bmatrix} 0 &0\\ 0 &0 \\ 0 &0 \end{bmatrix}$

And (-A) + A = $\begin{bmatrix} -a &-b \\ c &-d \\ -e &f \end{bmatrix}$ + $\begin{bmatrix} a &b \\ -c &d \\ e &-f \end{bmatrix}$

= $\begin{bmatrix} 0 &0\\ 0 &0 \\ 0 &0 \end{bmatrix}$
So, we get A + (-A) = 0 = (-A) + A
It is clear that -A is the inverse of A. This is one of the Matrix Inverse Properties.

## Distributive Law of Matrix Addition

Distributive law of Matrix Addition is stated as follows:

If A and B be two matrices of same order and c be any scalar, then C(A + B) = cA + cB

Proof:

Let A2x4 = $\begin{bmatrix} 1 &3 &5 &7 \\ 2 &-4 &8 &6 \end{bmatrix}$, B2x4 = $\begin{bmatrix} -2 &4 &6 &1 \\ 7&5 &2 &8 \end{bmatrix}$ and c be any scalar(real number).

A + B = $\begin{bmatrix} 1 &3 &5 &7 \\ 2 &-4 &8 &6 \end{bmatrix}$ + $\begin{bmatrix} -2 &4 &6 &1 \\ 7 &5 &2 &8 \end{bmatrix}$

= $\begin{bmatrix} -1 &7 &11 &8 \\ 9 &1 &10 &14 \end{bmatrix}$

c(A + B) = c $\begin{bmatrix} -1 &7 &11 &8 \\ 9 &1 &10 &14 \end{bmatrix}$

= $\begin{bmatrix} -c &7c &11c &8c \\ 9c &c &10c &14c \end{bmatrix}$

Again cA + cB = c $\begin{bmatrix} 1 &3 &5 &7 \\ 2 &-4 &8 &6 \end{bmatrix}$ + c $\begin{bmatrix} -2 &4 &6 &1 \\ 7&5 &2 &8 \end{bmatrix}$

= $\begin{bmatrix} c &3c &5c &7c \\ 2c &-4c &8c &c6 \end{bmatrix}$ + $\begin{bmatrix} -2c &4c &6c &1c \\ 7c&5c &2c &8c \end{bmatrix}$

= $\begin{bmatrix} -c &7c &11c &8c \\ 9c &c &10c &14c \end{bmatrix}$

= c(A + B) = cA + cB.

## Distributive Law of Scalar Addition

Distributive law of scalar addition is stated as follows:

If 4I3x3 is any scalar and A3x3 is any matrix, then 4 I3x3+ A = A + 4 I3x3.

Proof:
I3x3 be the identity matrix and A = $\begin{bmatrix} 2 &3 &4 \\ 5 &6 &7 \\ 8 &9 &0 \end{bmatrix}$

Then,
4 I3x3 + A = 4 $\begin{bmatrix} 1&0 &0 \\ 0 &1 &0\\ 0 &0 &1 \end{bmatrix}$ + $\begin{bmatrix} 2 &3 &4 \\ 5 &6 &7 \\ 8 &9 &0 \end{bmatrix}$

= $\begin{bmatrix} 4&0 &0 \\ 0 &4 &0\\ 0 &0 &4 \end{bmatrix}$ + $\begin{bmatrix} 2 &3 &4 \\ 5 &6 &7 \\ 8 &9 &0 \end{bmatrix}$

= $\begin{bmatrix} 6 &3 &4 \\ 5 &10 &7 \\ 8 &9 &4 \end{bmatrix}$ ......................................................................(i)

Again A + 4 I3x3 = $\begin{bmatrix} 2 &3 &4 \\ 5 &6 &7 \\ 8 &9 &0 \end{bmatrix}$ + 4 $\begin{bmatrix} 1&0 &0 \\ 0 &1 &0\\ 0 &0 &1 \end{bmatrix}$

= $\begin{bmatrix} 2 &3 &4 \\ 5 &6 &7 \\ 8 &9 &0 \end{bmatrix}$ + $\begin{bmatrix} 4&0 &0 \\ 0 &4 &0\\ 0 &0 &4 \end{bmatrix}$

= $\begin{bmatrix} 6 &3 &4 \\ 5 &10 &7 \\ 8 &9 &4 \end{bmatrix}$ ...................................................................(ii)

From (i) and (ii), we get 4 I3x3 + A = A + 4 I3x3

## Associative Law of Scalar Multiplication

Associative law for scalar multiplication is stated as follows:

If A is any matrix and c and d are two scalars then, (cd)A = c(dA).

Proof:
For this, let A3x2 = $\begin{bmatrix} - 4 &6 \\ 7&1 \\ 2&2 \end{bmatrix}$ is a matrix and 3, 2 are two scalars then,

(3 x 2) $\begin{bmatrix} - 4 &6 \\ 7&1 \\ 2&2 \end{bmatrix}$ = 3 x $\left (2\times \begin{bmatrix} -4 &6 \\ 7&1 \\ 2&2 \end{bmatrix} \right )$

$\Rightarrow$ 6 x $\begin{bmatrix} - 4 &6 \\ 7&1 \\ 2&2 \end{bmatrix}$ = 3 x $\begin{bmatrix} - 8 &12 \\ 14&2 \\ 4&4 \end{bmatrix}$

$\Rightarrow$ $\begin{bmatrix} - 24 &36 \\ 42&6 \\ 12&12 \end{bmatrix}$ = $\begin{bmatrix} - 24 &36 \\ 42&6 \\ 12&12 \end{bmatrix}$

$\Rightarrow$ (cd)A = c(dA)
This is one of the matrix multiplication properties.

## Rule for Multiplication by 1

In matrix multiplication, 1(identity matrix In) is the identity element of matrix multiplication, the rule for multiplication by 1 is stated as follows:

If A is any matrix and In and identity matrix then, InA= A = AIn

Proof:

A = $\begin{bmatrix} -2 &3 &4 \\ 5 &7 &1 \\ -2 &4 &5 \end{bmatrix}$

I3 A = $\begin{bmatrix} 1 &0&0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$ x $\begin{bmatrix} -2 &3 &4 \\ 5 &7 &1 \\ -2 &4 &5 \end{bmatrix}$

= $\begin{bmatrix} -2 &3 &4 \\ 5 &7 &1 \\ -2 &4 &5 \end{bmatrix}$

= A ...............................................................(a)

AIn = $\begin{bmatrix} -2 &3 &4 \\ 5 &7 &1 \\ -2 &4 &5 \end{bmatrix}$ x $\begin{bmatrix} 1 &0&0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$

= $\begin{bmatrix} -2 &3 &4 \\ 5 &7 &1 \\ -2 &4 &5 \end{bmatrix}$
= A ..............................................................(b)
From (a) and (b), it is clear that InA= A = AIn.