Addition of matrices:

Two matrices can be added only when they are of the same order

If two matrices of A and B are of the same order, they are said to be conformable for addition.

If A and B are matrices of order m × n, then their sum will also be a matrix of the same order and in addition, corresponding elements of A and B get added.

So if $A=\left[a_{i j}\right]_{m \times n}, B=\left[b_{i j}\right]_{m \times n}$ Then, $A+B=\left[a_{i j}+b_{i j}\right]_{m \times n}$ for all $i, j$ eg

$
\begin{aligned}
A & =\left[\begin{array}{lll}
10 & 20 & 30 \\
20 & 30 & 40 \\
30 & 40 & 50
\end{array}\right], \quad B=\left[\begin{array}{lll}
50 & 40 & 30 \\
40 & 30 & 20 \\
30 & 20 & 10
\end{array}\right] \\
A+B & =\left[\begin{array}{lll}
10+50 & 20+40 & 30+30 \\
20+40 & 30+30 & 40+20 \\
30+30 & 40+20 & 50+10
\end{array}\right]=\left[\begin{array}{lll}
60 & 60 & 60 \\
60 & 60 & 60 \\
60 & 60 & 60
\end{array}\right]
\end{aligned}
$

Subtraction of matrices:

Two matrices can be subtracted only when they are of the same order. If A and B are matrices of order m × n then their difference will also be a matrix of the same order and in subtraction, corresponding elements of A and B get subtracted. So if 

$A=\left[a_{i j}\right]_{m \times n}, B=\left[b_{i j}\right]_{m \times n}$ Then, $A-B=\left[a_{i j}-b_{i j}\right]_{m \times n}$ for all $i, j$ eg

$
\begin{aligned}
\mathrm{A} & =\left[\begin{array}{lll}
10 & 20 & 30 \\
20 & 30 & 40 \\
30 & 40 & 50
\end{array}\right], \quad \mathrm{B}=\left[\begin{array}{lll}
50 & 40 & 30 \\
40 & 30 & 20 \\
30 & 20 & 10
\end{array}\right] \\
\mathrm{A}-\mathrm{B} & =\left[\begin{array}{lll}
10-50 & 20-40 & 30-30 \\
20-40 & 30-30 & 40-20 \\
30-30 & 40-20 & 50-10
\end{array}\right]=\left[\begin{array}{ccc}
-40 & -20 & 0 \\
-20 & 0 & 20 \\
0 & 20 & 40
\end{array}\right]
\end{aligned}
$
 

Properties of matrix addition:

1. Matrix addition is commutative, $\mathrm{A}+\mathrm{B}=\mathrm{B}+\mathrm{A}$
2. Matrix addition is associative, $A+(B+C)=(A+B)+C$
3. Additive identity exist, means there exist a matrix $O$ (null matrix) such that $A+O=A=O+A$ (Here $O$ has same order as A )
4. Existence of additive inverse means there exists a matrix $B$ such that $A+B=O=B+A$
5. Cancellation property:

If $A+B=A+C$ then $B=C$
If $\mathrm{A}+\mathrm{C}=\mathrm{B}+\mathrm{C}$ then $\mathrm{A}=\mathrm{B}$
Note: all matrix taken in above property explanation has same order which is $\mathrm{m} \times \mathrm{n}$.

Scalar multiplication:

Let k be any scalar number, and $A=\left[a_{i j}\right]_{m \times n}$  be a matrix. Then the matrix is obtained by multiplying every element A by a scalar k and denoted as kA.

$\begin{aligned} & k A=\left[k a_{i j}\right]_{m \times n} \\ & \qquad \mathrm{~A}=\left[\begin{array}{ll}2 & 6 \\ 3 & 7 \\ 5 & 8\end{array}\right] \text { then, } 3 \mathrm{~A}=\left[\begin{array}{ll}3 \times 2 & 3 \times 6 \\ 3 \times 3 & 3 \times 7 \\ 3 \times 5 & 3 \times 8\end{array}\right]=\left[\begin{array}{cc}6 & 18 \\ 9 & 21 \\ 15 & 24\end{array}\right]\end{aligned}$

Properties of scalar multiplication: 

If $A$ and $B$ are two matrices and $k, 1$ are scalar then
i) $k(A+B)=k A+k B$
ii) $k l(A)=k(I A)=l(k A)$
iii) $(k+1) A=k A+I A$
iv) $(-k) A=-(k A)=k(-A)$
v) $1 \mathrm{~A}=\mathrm{A},(-1) \mathrm{A}=-\mathrm{A}$

Note: $A$ and $B$ have the same order $m \times n$.