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Let $A = \begin{pmatrix} 0 & -3 \\ 2 & -4 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$. Find $A\mathbf{v}$.
Row 1, column 1: Hint: Row 1 of $A$ is $\begin{pmatrix} 0 & -3 \end{pmatrix}$. Column 1 of $\mathbf{v}$ is $\begin{pmatrix} 4 \\ 4 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a0}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b0}}\hspace{40px}}~} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a1}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b1}}\hspace{40px}}~}$ $0 \times 4 + -3 \times 4$
$= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p0}}\hspace{54px}}~} + \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p1}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0R}}\hspace{54px}}~}$ $= 0 + -12 = -12$
Row 2, column 1: Hint: Row 2 of $A$ is $\begin{pmatrix} 2 & -4 \end{pmatrix}$. Column 1 of $\mathbf{v}$ is $\begin{pmatrix} 4 \\ 4 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a0}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b0}}\hspace{40px}}~} + \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a1}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b1}}\hspace{40px}}~}$ $2 \times 4 + -4 \times 4$
$= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p0}}\hspace{54px}}~} + \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p1}}\hspace{54px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0R}}\hspace{54px}}~}$ $= 8 + -16 = -8$
Therefore,
$A\mathbf{v} = \begin{pmatrix} \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c0}}\hspace{54px}}~} \\ \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c0}}\hspace{54px}}~} \end{pmatrix}$ $A\mathbf{v} = \begin{pmatrix} -12 \\ -8 \end{pmatrix}$