Solve the problem.
-Let $\mathrm { A } = \left[ \begin{array} { r r } - 3 & 5 \ 0 & 2 \end{array} \right]$. Find $4 \mathrm {~A}$.
A)
$\left[ \begin{array} { r r } - 12 & 20 \ 0 & 8 \end{array} \right]$
B)
$\left[ \begin{array} { r r } - 12 & 20 \ 0 & 2 \end{array} \right]$
C)
$\left[ \begin{array} { r } - 125 \ 02 \end{array} \right]$
D)
$\left[ \begin{array} { l l } 1 & 9 \ 4 & 6 \end{array} \right]$

Question

Quizplus · Accepted Answer

To multiply a matrix by a scalar, we simply multiply each element of the matrix by the scalar. In this case, we want to find $4 \mathrm{~A}$, so we need to multiply each element of $\mathrm{A}$ by $4$.
$$
4 \mathrm {~A} = 4 \left[ \begin{array} { r r } - 3 & 5 \ 0 & 2 \end{array} \right] = \left[ \begin{array} { r r } 4 \cdot (-3) & 4 \cdot 5 \ 4 \cdot 0 & 4 \cdot 2 \end{array} \right] = \left[ \begin{array} { r r } -12 & 20 \ 0 & 8 \end{array} \right]
$$
Therefore, the answer is choice A.

Solve the Problem A=[−3502]\mathrm { A } = \left[ \begin{array} { r r } - 3 & 5 \\ 0 & 2 \end{array} \right]A=[−30​52​]

Solve the Problem $\mathrm { A } = \left[ \begin{array} { r r } - 3 & 5 \\ 0 & 2 \end{array} \right]$