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