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Let $A = \begin{pmatrix} -4 & 4 \\ -2 & -1 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$. Find $A\mathbf{v}$.
Row 1, column 1: Hint: Row 1 of $A$ is $\begin{pmatrix} -4 & 4 \end{pmatrix}$. Column 1 of $\mathbf{v}$ is $\begin{pmatrix} 3 \\ 3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a0}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b0}}\hspace{40px}}~} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a1}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b1}}\hspace{40px}}~}$ $-4 \times 3 + 4 \times 3$
$= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p0}}\hspace{54px}}~} + \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p1}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0R}}\hspace{54px}}~}$ $= -12 + 12 = 0$
Row 2, column 1: Hint: Row 2 of $A$ is $\begin{pmatrix} -2 & -1 \end{pmatrix}$. Column 1 of $\mathbf{v}$ is $\begin{pmatrix} 3 \\ 3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a0}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b0}}\hspace{40px}}~} + \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a1}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b1}}\hspace{40px}}~}$ $-2 \times 3 + -1 \times 3$
$= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p0}}\hspace{54px}}~} + \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p1}}\hspace{54px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0R}}\hspace{54px}}~}$ $= -6 + -3 = -9$
Therefore,
$A\mathbf{v} = \begin{pmatrix} \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c0}}\hspace{54px}}~} \\ \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c0}}\hspace{54px}}~} \end{pmatrix}$ $A\mathbf{v} = \begin{pmatrix} 0 \\ -9 \end{pmatrix}$