We proceed to develop the algebra of addition of matrices.

Two matrices A and B are said to be conformable for addition if they have the same order (i.e., same number of rows and columns).

If A  = (aij)m, n and B = (bij)m, n then their sum A + B is the matrix C = (cij)m,n where cij = aij + bij, i = 1, 2, 3, ...... , m, j = 1, 2, 3, ...., n.

For example:

If A = $$\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$ and B = $$\begin{bmatrix} b_{11} & b_{12} & b_{13}\\ b_{21} & b_{22} & b_{23}\\ b_{31} & b_{32} & b_{33} \end{bmatrix}$$, then

A + B = $$\begin{bmatrix} a_{11} + b_{11} & a_{12} + b_{12} & a_{13} + b_{13}\\ a_{21} + b_{21} & a_{22} + b_{22} & a_{23} + b_{23}\\ a_{31} + b_{31} & a_{32} + b_{32} & a_{33} + b_{33} \end{bmatrix}$$ = C

Note: If A and B be matrices of different orders, then A + B is not defined.

1. If A = $$\begin{bmatrix} 2 & 5\\ -1 & 3 \end{bmatrix}$$ and B = $$\begin{bmatrix} 1 & 4\\ 3 & 7 \end{bmatrix}$$, then

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

= $$\begin{bmatrix} 3 & 9\\ 2 & 10 \end{bmatrix}$$

2. If A = $$\begin{bmatrix} -1 & 2 & 3\\ 2 & -3 & 1\\ 3 & 1 & -2 \end{bmatrix}$$, B = $$\begin{bmatrix} 3 & -1 & 2\\ 1 & 0 & 3\\ 2 & -1 & 0 \end{bmatrix}$$ and M = $$\begin{bmatrix} 5 & 2\\ 1 & 4 \end{bmatrix}$$, then

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

= $$\begin{bmatrix} -1 + 3 & 2 + (- 1) & 3 + 2\\ 2 + 1 & -3 + 0 & 1 + 3\\ 3 + 2 & 1 + (-1) & -2 + 0 \end{bmatrix}$$

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

A + M is not defined since the order of matrix M is not equal to the order of matrix A.

B + M is also not defined since the order of matrix M is not equal to the order of matrix B.

Note: Let A and B are m × n matrices and c, d are scalars. Then the following results are obvious.

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

For Example:

If A = $$\begin{bmatrix} 2 & 5\\ 3 & 1 \end{bmatrix}$$ and B =  $$\begin{bmatrix} 3 & 2\\ 0 & 7 \end{bmatrix}$$ are m × n matrices and 5 is scalar. Then

$5\left (\begin{bmatrix} 2 & 5\\ 3 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 2\\ 0 & 7 \end{bmatrix} \right ) = 5\begin{bmatrix} 2 & 5\\ 3 & 1 \end{bmatrix} + 5 \begin{bmatrix} 3 & 2\\ 0 & 7 \end{bmatrix}$

II. (c + d)A = cA + dA.

For Example:

If A = $$\begin{bmatrix} 2 & 5\\ 3 & 1 \end{bmatrix}$$ be m × n matrix and 5, 3 are scalars. Then

$\left (5 + 3\right )\begin{bmatrix} 2 & 5\\ 3 & 1 \end{bmatrix} = 5\begin{bmatrix} 2 & 5\\ 3 & 1 \end{bmatrix} + 3\begin{bmatrix} 2 & 5\\ 3 & 1 \end{bmatrix}$

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