### Theory:

We can multiply the elements of the given matrix $$A$$ by a non-zero number $$k$$ to obtain a new matrix $$kA$$ whose elements are multiplied by $$k$$.

The matrix $$kA$$ is called scalar multiplication of $$A$$.

If $$A = (a_{ij})_{m×n}$$,  then, $$kA = (ka_{ij})_{m×n}$$ for all $$i = 1, 2,...m$$, and such that $$j = 1, 2, ….n$$
Example:
Determine 3$$A + B$$, if $$A =\begin{bmatrix} 2 & 4 & 6\\ 7 & 5 & -4\\ -2 & 1 & 7 \end{bmatrix}, B = \begin{bmatrix} 2 & 4 & 6\\ 7 & 5 & 3\\ 7 & 1 & 7 \end{bmatrix}$$

Both the matrices $$A$$ and $$B$$ have same orders as $$3 × 3$$, so 3$$A + B$$ is defined.

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

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

$$= \begin{bmatrix} 6 & 12 & 18 \\ 15 & 12 & -12\\ -6 & 3 & 21 \end{bmatrix} + \begin{bmatrix} 2 & 4 & 6\\ 7 & 5 & 3\\ 7 & 1 & 7 \end{bmatrix}$$

Now we add the two matrices.

$$= \begin{bmatrix} 8 & 16 & 24 \\ 22 & 17 & -9\\ 1 & 4 & 28 \end{bmatrix}$$