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Let $A = \begin{pmatrix} -1 & -2 \\ 8 & -1 \\ -5 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 7 & 3 \\ -3 & -3 \end{pmatrix}$. Find $AB$.
Row 1, column 1: Hint: Row 1 of $A$ is $\begin{pmatrix} -1 & -2 \end{pmatrix}$. Column 1 of $B$ is $\begin{pmatrix} 7 \\ -3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a0}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b0}}\hspace{40px}}~} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a1}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b1}}\hspace{40px}}~}$ $-1 \times 7 + -2 \times -3$
$= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p0}}\hspace{54px}}~} + \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p1}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0R}}\hspace{54px}}~}$ $= -7 + 6 = -1$
Row 1, column 2: Hint: Row 1 of $A$ is $\begin{pmatrix} -1 & -2 \end{pmatrix}$. Column 2 of $B$ is $\begin{pmatrix} 3 \\ -3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1a0}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1b0}}\hspace{40px}}~} + \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1a1}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1b1}}\hspace{40px}}~}$ $-1 \times 3 + -2 \times -3$
$= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1p0}}\hspace{54px}}~} + \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1p1}}\hspace{54px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1R}}\hspace{54px}}~}$ $= -3 + 6 = 3$
Row 2, column 1: Hint: Row 2 of $A$ is $\begin{pmatrix} 8 & -1 \end{pmatrix}$. Column 1 of $B$ is $\begin{pmatrix} 7 \\ -3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a0}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b0}}\hspace{40px}}~} + \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a1}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b1}}\hspace{40px}}~}$ $8 \times 7 + -1 \times -3$
$= \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p0}}\hspace{54px}}~} + \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p1}}\hspace{54px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0R}}\hspace{54px}}~}$ $= 56 + 3 = 59$
Row 2, column 2: Hint: Row 2 of $A$ is $\begin{pmatrix} 8 & -1 \end{pmatrix}$. Column 2 of $B$ is $\begin{pmatrix} 3 \\ -3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1a0}}\hspace{40px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1b0}}\hspace{40px}}~} + \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1a1}}\hspace{40px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1b1}}\hspace{40px}}~}$ $8 \times 3 + -1 \times -3$
$= \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1p0}}\hspace{54px}}~} + \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1p1}}\hspace{54px}}~} = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1R}}\hspace{54px}}~}$ $= 24 + 3 = 27$
Row 3, column 1: Hint: Row 3 of $A$ is $\begin{pmatrix} -5 & 4 \end{pmatrix}$. Column 1 of $B$ is $\begin{pmatrix} 7 \\ -3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0a0}}\hspace{40px}}~} \times \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0b0}}\hspace{40px}}~} + \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0a1}}\hspace{40px}}~} \times \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0b1}}\hspace{40px}}~}$ $-5 \times 7 + 4 \times -3$
$= \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0p0}}\hspace{54px}}~} + \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0p1}}\hspace{54px}}~} = \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0R}}\hspace{54px}}~}$ $= -35 + -12 = -47$
Row 3, column 2: Hint: Row 3 of $A$ is $\begin{pmatrix} -5 & 4 \end{pmatrix}$. Column 2 of $B$ is $\begin{pmatrix} 3 \\ -3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step11}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1a0}}\hspace{40px}}~} \times \class{inputBox step11}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1b0}}\hspace{40px}}~} + \class{inputBox step11}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1a1}}\hspace{40px}}~} \times \class{inputBox step11}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1b1}}\hspace{40px}}~}$ $-5 \times 3 + 4 \times -3$
$= \class{inputBox step12}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1p0}}\hspace{54px}}~} + \class{inputBox step12}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1p1}}\hspace{54px}}~} = \class{inputBox step12}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1R}}\hspace{54px}}~}$ $= -15 + -12 = -27$
Therefore,
$AB = \begin{pmatrix} \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c0}}\hspace{54px}}~} & \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c1}}\hspace{54px}}~} \\ \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c0}}\hspace{54px}}~} & \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c1}}\hspace{54px}}~} \\ \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr2c0}}\hspace{54px}}~} & \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr2c1}}\hspace{54px}}~} \end{pmatrix}$ $AB = \begin{pmatrix} -1 & 3 \\ 59 & 27 \\ -47 & -27 \end{pmatrix}$