Find the matrix product when possible.
-$$\left[ \begin{array} { r r }
3 & - 1 \
6 & 0
\end{array} \right] \left[ \begin{array} { r r }
0 & - 1 \
2 & 6
\end{array} \right]$$
A)
$\left[ \begin{array} { r r } 0 & 1 \ 12 & 0 \end{array} \right]$

B)
$\left[ \begin{array} { c c } - 9 & - 2 \ - 6 & 0 \end{array} \right]$

C)
$\left[ \begin{array} { r r } - 2 & - 9 \ 0 & - 6 \end{array} \right]$

D)
$\left[ \begin{array} { c c } - 6 & 0 \ 42 & - 2 \end{array} \right]$

Question

Find the matrix product when possible.
-$$\left[ \begin{array} { r r } 
3 & - 1 \
6 & 0
\end{array} \right] \left[ \begin{array} { r r } 
0 & - 1 \
2 & 6
\end{array} \right]$$
A)
$\left[ \begin{array} { r r } 0 & 1 \ 12 & 0 \end{array} \right]$

B)
$\left[ \begin{array} { c c } - 9 & - 2 \ - 6 & 0 \end{array} \right]$

C)
$\left[ \begin{array} { r r } - 2 & - 9 \ 0 & - 6 \end{array} \right]$

D)
$\left[ \begin{array} { c c } - 6 & 0 \ 42 & - 2 \end{array} \right]$

Quizplus · Accepted Answer

To find the product of two matrices, multiply the elements of the rows of the first matrix by the corresponding elements of the columns in the second matrix and add them up. For the given matrices, the product is calculated as follows:- For the element in the first row, first column: $3*0 + (-1)*2 = 0 - 2 = -2$- For the element in the first row, second column: $3*(-1) + (-1)*6 = -3 - 6 = -9$- For the element in the second row, first column: $6*0 + 0*2 = 0 + 0 = 0$- For the element in the second row, second column: $6*(-1) + 0*6 = -6 + 0 = -6$Thus, the product matrix is $\left[ \begin{array} { r r } - 2 & - 9 \ 0 & - 6 \end{array} \right]$.

Find the Matrix Product When Possible [3−160][0−126]\left[ \begin{array} { r r } 3 & - 1 \\6 & 0\end{array} \right] \left[ \begin{array} { r r } 0 & - 1 \\2 & 6\end{array} \right][36​−10​][02​−16​]

Find the Matrix Product When Possible $\left[ \begin{array} { r r } 3 & - 1 \\6 & 0\end{array} \right] \left[ \begin{array} { r r } 0 & - 1 \\2 & 6\end{array} \right]$