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Let $A = \begin{pmatrix} 3 & -3 \\ 4 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} -4 & -4 \\ -3 & 0 \end{pmatrix}$. Find $AB$.
Row 1, column 1: Hint: Row 1 of $A$ is $\begin{pmatrix} 3 & -3 \end{pmatrix}$. Column 1 of $B$ is $\begin{pmatrix} -4 \\ -3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a0}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b0}}\hspace{40px}}~} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a1}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b1}}\hspace{40px}}~}$ $[3] \times [-4] + [-3] \times [-3]$
$= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p0}}\hspace{54px}}~} + \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p1}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0R}}\hspace{54px}}~}$ $= [-12] + [9] = [-3]$
Row 1, column 2: Hint: Row 1 of $A$ is $\begin{pmatrix} 3 & -3 \end{pmatrix}$. Column 2 of $B$ is $\begin{pmatrix} -4 \\ 0 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1a0}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1b0}}\hspace{40px}}~} + \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1a1}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1b1}}\hspace{40px}}~}$ $[3] \times [-4] + [-3] \times [0]$
$= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1p0}}\hspace{54px}}~} + \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1p1}}\hspace{54px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1R}}\hspace{54px}}~}$ $= [-12] + [0] = [-12]$
Row 2, column 1: Hint: Row 2 of $A$ is $\begin{pmatrix} 4 & 1 \end{pmatrix}$. Column 1 of $B$ is $\begin{pmatrix} -4 \\ -3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a0}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b0}}\hspace{40px}}~} + \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a1}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b1}}\hspace{40px}}~}$ $[4] \times [-4] + [1] \times [-3]$
$= \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p0}}\hspace{54px}}~} + \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p1}}\hspace{54px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0R}}\hspace{54px}}~}$ $= [-16] + [-3] = [-19]$
Row 2, column 2: Hint: Row 2 of $A$ is $\begin{pmatrix} 4 & 1 \end{pmatrix}$. Column 2 of $B$ is $\begin{pmatrix} -4 \\ 0 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1a0}}\hspace{40px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1b0}}\hspace{40px}}~} + \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1a1}}\hspace{40px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1b1}}\hspace{40px}}~}$ $[4] \times [-4] + [1] \times [0]$
$= \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1p0}}\hspace{54px}}~} + \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1p1}}\hspace{54px}}~} = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1R}}\hspace{54px}}~}$ $= [-16] + [0] = [-16]$
Therefore,
$AB = \begin{pmatrix} \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c0}}\hspace{54px}}~} & \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c1}}\hspace{54px}}~} \\ \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c0}}\hspace{54px}}~} & \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c1}}\hspace{54px}}~} \end{pmatrix}$ $AB = \begin{pmatrix} [-3] & [-12] \\ [-19] & [-16] \end{pmatrix}$