$k \times \big[\,e_{ij}\,\big] \,=\, \big[k \times e_{ij}\big]$
The multiplication of a matrix by a scalar is equal to the multiplication of each entry (or element) by the scalar in the matrix, is called the scalar multiplication rule of the matrices.
Let $k$ be a constant and the matrix $M$ represents a matrix of the order $m \times n$.
$M$ $\,=\,$ $\begin{bmatrix} e_{11} & e_{12} & e_{13} & \cdots & e_{1n}\\ e_{21} & e_{22} & e_{23} & \cdots & e_{2n}\\ e_{31} & e_{32} & e_{33} & \cdots & e_{3n}\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ e_{m1} & e_{m2} & e_{m3} & \cdots & e_{mn} \end{bmatrix}$
Let’s assume that the matrix $M$ is multiplied by the scalar $k$.
$\implies$ $k \times M$ $\,=\,$ $k \times \begin{bmatrix} e_{11} & e_{12} & e_{13} & \cdots & e_{1n}\\ e_{21} & e_{22} & e_{23} & \cdots & e_{2n}\\ e_{31} & e_{32} & e_{33} & \cdots & e_{3n}\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ e_{m1} & e_{m2} & e_{m3} & \cdots & e_{mn} \end{bmatrix}$
The product of a scalar and a matrix is equal to the product of each element and scalar in a matrix.
$\implies$ $k \times M$ $\,=\,$ $\begin{bmatrix} k \times e_{11} & k \times e_{12} & k \times e_{13} & \cdots & k \times e_{1n}\\ k \times e_{21} & k \times e_{22} & k \times e_{23} & \cdots & k \times e_{2n}\\ k \times e_{31} & k \times e_{32} & k \times e_{33} & \cdots & k \times e_{3n}\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ k \times e_{m1} & k \times e_{m2} & k \times e_{m3} & \cdots & k \times e_{mn} \end{bmatrix}$
This mathematical property of matrices is called the scalar multiplication of the matrices. It is used as a formula to multiply any matrix by any scalar.
If the matrix $M$ is simply expressed as $\big[e_{ij}\big]$, then the scalar multiplication rule can be simply written as follows.
$\implies$ $k \big[e_{ij}\big] \,=\, \big[k e_{ij}\big]$
Here, the letters $i$ and $j$ represent “the number of the row” and “the number of the column”.
Let’s learn how to use the scalar multiplication rule of the matrices from some understandable examples.
$(1).\,\,\,$ $4$ $\times$ $\begin{bmatrix} 1 & 7 \\ -2 & 6 \end{bmatrix}$
In this example, the matrix of the order $2$ is multiplied by a scalar $4$. It can be evaluated by multiplying each entry in the matrix by the scalar $4$.
$\implies$ $4$ $\times$ $\begin{bmatrix} 1 & 7 \\ -2 & 6 \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 4 \times 1 & 4 \times 7 \\ 4 \times (-2) & 4 \times 6 \end{bmatrix}$
$\,\,\,\therefore\,\,\,\,\,\,$ $4$ $\times$ $\begin{bmatrix} 1 & 7 \\ -2 & 6 \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 4 & 28 \\ -8 & 24 \end{bmatrix}$
$(2).\,\,\,$ $-1$ $\times$ $\begin{bmatrix} 2 & 0 & 3 & 5 \\ 5 & 1 & 8 & 4 \\ 7 & 3 & 2 & 6 \end{bmatrix}$
In this case, a matrix of the order $3 \times 4$ is multiplied by the scalar $-1$. So, multiply each entry in this matrix by the number -1.
$\implies$ $-1$ $\times$ $\begin{bmatrix} 2 & 0 & 3 & 5 \\ 5 & 1 & 8 & 4 \\ 7 & 3 & 2 & 6 \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} (-1) \times 2 & (-1) \times 0 & (-1) \times 3 & (-1) \times 5 \\ (-1) \times 5 & (-1) \times 1 & (-1) \times 8 & (-1) \times 4 \\ (-1) \times 7 & (-1) \times 3 & (-1) \times 2 & (-1) \times 6 \end{bmatrix}$
$\,\,\,\therefore\,\,\,\,\,\,$ $-1$ $\times$ $\begin{bmatrix} 2 & 0 & 3 & 5 \\ 5 & 1 & 8 & 4 \\ 7 & 3 & 2 & 6 \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} -2 & 0 & -3 & -5 \\ -5 & -1 & -8 & -4 \\ -7 & -3 & -2 & -6 \end{bmatrix}$
$(3).\,\,\,$ $6$ $\times$ $\begin{bmatrix} -9 & 5 & -2 \\ 5 & 8 & -3 \\ 6 & -1 & 0 \end{bmatrix}$
Use the scalar multiplication of matrices formula for multiplying the square matrix of the order $3$ by the scalar $6$.
$\implies$ $6$ $\times$ $\begin{bmatrix} -9 & 5 & -2 \\ 5 & 8 & -3 \\ 6 & -1 & 0 \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} 6 \times (-9) & 6 \times 5 & 6 \times (-2) \\ 6 \times 5 & 6 \times 8 & 6 \times (-3) \\ 6 \times 6 & 6 \times (-1) & 6 \times 0 \end{bmatrix}$
$\,\,\,\therefore\,\,\,\,\,\,$ $6$ $\times$ $\begin{bmatrix} -9 & 5 & -2 \\ 5 & 8 & -3 \\ 6 & -1 & 0 \end{bmatrix}$ $\,=\,$ $\begin{bmatrix} -54 & 30 & -12 \\ 30 & 48 & -18 \\ 36 & -6 & 0 \end{bmatrix}$
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