Solve the problem.
-Given $A = \left[ \begin{array} { l l } 2 & - 1 \ 3 & - 2 \end{array} \right]$ and $B = \left[ \begin{array} { r r } 0 & - 1 \ 4 & 3 \end{array} \right]$, find $A B$ and $B A$.
A)
$\mathrm { AB } = \left[ \begin{array} { l l } 4 & 1 \ 8 & 3 \end{array} \right] ; \mathrm { BA } = \left[ \begin{array} { r r } 3 & - 2 \ - 1 & 2 \end{array} \right]$
B)
$\mathrm { AB } = \left[ \begin{array} { r r } - 4 & - 5 \ - 8 & - 9 \end{array} \right] ; \mathrm { BA } = \left[ \begin{array} { r r } - 3 & 2 \ 17 & - 10 \end{array} \right]$

C)
$\mathrm { AB } = \left[ \begin{array} { r r } - 3 & 2 \ 17 & - 10 \end{array} \right] ; \mathrm { BA } = \left[ \begin{array} { r r } - 4 & - 5 \ - 8 & - 9 \end{array} \right]$
D)

$\mathrm { AB } = \left[ \begin{array} { r r } 0 & 1 \ 12 & - 6 \end{array} \right] ; \mathrm { BA } = \left[ \begin{array} { r r } 0 & 1 \ 12 & - 6 \end{array} \right]$

Question

Solve the problem.
-Given $A = \left[ \begin{array} { l l } 2 & - 1 \ 3 & - 2 \end{array} \right]$ and $B = \left[ \begin{array} { r r } 0 & - 1 \ 4 & 3 \end{array} \right]$, find $A B$ and $B A$.
A)
$\mathrm { AB } = \left[ \begin{array} { l l } 4 & 1 \ 8 & 3 \end{array} \right] ; \mathrm { BA } = \left[ \begin{array} { r r } 3 & - 2 \ - 1 & 2 \end{array} \right]$
B)
$\mathrm { AB } = \left[ \begin{array} { r r } - 4 & - 5 \ - 8 & - 9 \end{array} \right] ; \mathrm { BA } = \left[ \begin{array} { r r } - 3 & 2 \ 17 & - 10 \end{array} \right]$

C)
$\mathrm { AB } = \left[ \begin{array} { r r } - 3 & 2 \ 17 & - 10 \end{array} \right] ; \mathrm { BA } = \left[ \begin{array} { r r } - 4 & - 5 \ - 8 & - 9 \end{array} \right]$
D)

$\mathrm { AB } = \left[ \begin{array} { r r } 0 & 1 \ 12 & - 6 \end{array} \right] ; \mathrm { BA } = \left[ \begin{array} { r r } 0 & 1 \ 12 & - 6 \end{array} \right]$

Quizplus · Accepted Answer

To find $AB$, multiply each row of $A$ by each column of $B$ and sum the products: $AB = \left[ \begin{array} { r r } (2*0)+(-1*4) & (2*-1)+(-1*3) \ (3*0)+(-2*4) & (3*-1)+(-2*3) \end{array} \right] = \left[ \begin{array} { r r } -4 & -5 \ -8 & -9 \end{array} \right]$. For $BA$, multiply each row of $B$ by each column of $A$: $BA = \left[ \begin{array} { r r } (0*2)+(-1*3) & (0*-1)+(-1*-2) \ (4*2)+(3*3) & (4*-1)+(3*-2) \end{array} \right] = \left[ \begin{array} { r r } -3 & 2 \ 17 & -10 \end{array} \right]$.

Solve the Problem A=[2−13−2]A = \left[ \begin{array} { l l } 2 & - 1 \\ 3 & - 2 \end{array} \right]A=[23​−1−2​]

Solve the Problem $A = \left[ \begin{array} { l l } 2 & - 1 \\ 3 & - 2 \end{array} \right]$