Formulaic Maths Problems - get another problem: level 1, level 2, level 3 - link to this problem
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Let $A = \begin{pmatrix} 3 & -1 & -5 \\ 6 & 0 & -5 \\ 0 & -7 & -8 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix}$. Find $A\mathbf{v}$.
Row 1, column 1: Hint: Row 1 of $A$ is $\begin{pmatrix} 3 & -1 & -5 \end{pmatrix}$. Column 1 of $\mathbf{v}$ is $\begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a0}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b0}}\hspace{40px}}~} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a1}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b1}}\hspace{40px}}~} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a2}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b2}}\hspace{40px}}~}$ $3 \times 2 + -1 \times -4 + -5 \times 3$
$= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p0}}\hspace{54px}}~} + \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p1}}\hspace{54px}}~} + \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p2}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0R}}\hspace{54px}}~}$ $= 6 + 4 + -15 = -5$
Row 2, column 1: Hint: Row 2 of $A$ is $\begin{pmatrix} 6 & 0 & -5 \end{pmatrix}$. Column 1 of $\mathbf{v}$ is $\begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a0}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b0}}\hspace{40px}}~} + \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a1}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b1}}\hspace{40px}}~} + \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a2}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b2}}\hspace{40px}}~}$ $6 \times 2 + 0 \times -4 + -5 \times 3$
$= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p0}}\hspace{54px}}~} + \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p1}}\hspace{54px}}~} + \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p2}}\hspace{54px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0R}}\hspace{54px}}~}$ $= 12 + 0 + -15 = -3$
Row 3, column 1: Hint: Row 3 of $A$ is $\begin{pmatrix} 0 & -7 & -8 \end{pmatrix}$. Column 1 of $\mathbf{v}$ is $\begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0a0}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0b0}}\hspace{40px}}~} + \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0a1}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0b1}}\hspace{40px}}~} + \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0a2}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0b2}}\hspace{40px}}~}$ $0 \times 2 + -7 \times -4 + -8 \times 3$
$= \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0p0}}\hspace{54px}}~} + \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0p1}}\hspace{54px}}~} + \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0p2}}\hspace{54px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0R}}\hspace{54px}}~}$ $= 0 + 28 + -24 = 4$
Therefore,
$A\mathbf{v} = \begin{pmatrix} \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c0}}\hspace{54px}}~} \\ \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c0}}\hspace{54px}}~} \\ \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr2c0}}\hspace{54px}}~} \end{pmatrix}$ $A\mathbf{v} = \begin{pmatrix} -5 \\ -3 \\ 4 \end{pmatrix}$