Find the matrix product when possible.
-$$\left[ \begin{array} { r r }
- 1 & 3 \
2 & 2
\end{array} \right] \left[ \begin{array} { l l }
- 2 & 0 \
- 1 & 4
\end{array} \right]$$
A)
$\left[\begin{array}{rr}
2 & 0 \
-2 & 8
\end{array}\right]$
B)
$\left[\begin{array}{rr}
2 & -6 \
-1 & 5
\end{array}\right]$
C)
$\left[\begin{array}{rr}
-1 & 12 \
-6 & 8
\end{array}\right]$

D)
$\left[\begin{array}{rr}
12 & -1 \
8 & -6
\end{array}\right]$

Question

Find the matrix product when possible.
-$$\left[ \begin{array} { r r } 
- 1 & 3 \
2 & 2
\end{array} \right] \left[ \begin{array} { l l } 
- 2 & 0 \
- 1 & 4
\end{array} \right]$$
A)
$\left[\begin{array}{rr}
2 & 0 \
-2 & 8
\end{array}\right]$
B)
$\left[\begin{array}{rr}
2 & -6 \
-1 & 5
\end{array}\right]$
C)
$\left[\begin{array}{rr}
-1 & 12 \
-6 & 8
\end{array}\right]$

D)
$\left[\begin{array}{rr}
12 & -1 \
8 & -6
\end{array}\right]$

Quizplus · Accepted Answer

To find the product of two matrices, you multiply the rows of the first matrix by the columns of the second matrix. For this problem, the calculations are as follows:For the first element (top left), $(-1) \times (-2) + 3 \times (-1) = 2 - 3 = -1$.For the second element (top right), $(-1) \times 0 + 3 \times 4 = 0 + 12 = 12$.For the third element (bottom left), $2 \times (-2) + 2 \times (-1) = -4 - 2 = -6$.For the fourth element (bottom right), $2 \times 0 + 2 \times 4 = 0 + 8 = 8$.Thus, the product of the matrices is $\left[\begin{array}{rr} -1 & 12 \ -6 & 8 \end{array}\right]$.

Find the Matrix Product When Possible [−1322][−20−14]\left[ \begin{array} { r r } - 1 & 3 \\2 & 2\end{array} \right] \left[ \begin{array} { l l } - 2 & 0 \\- 1 & 4\end{array} \right][−12​32​][−2−1​04​]

Find the Matrix Product When Possible $\left[ \begin{array} { r r } - 1 & 3 \\2 & 2\end{array} \right] \left[ \begin{array} { l l } - 2 & 0 \\- 1 & 4\end{array} \right]$