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Let $A = \begin{pmatrix} -1 & -2 \\ 3 & -2 \end{pmatrix}$ and $B = \begin{pmatrix} -1 & 3 \\ -2 & 2 \end{pmatrix}$. Find $AB$.
Row 1, column 1: Hint: Row 1 of $A$ is $\begin{pmatrix} -1 & -2 \end{pmatrix}$. Column 1 of $B$ is $\begin{pmatrix} -1 \\ -2 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a0}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b0}}\hspace{40px}}~} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a1}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b1}}\hspace{40px}}~}$ $-1 \times -1 + -2 \times -2$
$= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p0}}\hspace{54px}}~} + \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p1}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0R}}\hspace{54px}}~}$ $= 1 + 4 = 5$
Row 1, column 2: Hint: Row 1 of $A$ is $\begin{pmatrix} -1 & -2 \end{pmatrix}$. Column 2 of $B$ is $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1a0}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1b0}}\hspace{40px}}~} + \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1a1}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1b1}}\hspace{40px}}~}$ $-1 \times 3 + -2 \times 2$
$= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1p0}}\hspace{54px}}~} + \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1p1}}\hspace{54px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1R}}\hspace{54px}}~}$ $= -3 + -4 = -7$
Row 2, column 1: Hint: Row 2 of $A$ is $\begin{pmatrix} 3 & -2 \end{pmatrix}$. Column 1 of $B$ is $\begin{pmatrix} -1 \\ -2 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a0}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b0}}\hspace{40px}}~} + \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a1}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b1}}\hspace{40px}}~}$ $3 \times -1 + -2 \times -2$
$= \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p0}}\hspace{54px}}~} + \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p1}}\hspace{54px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0R}}\hspace{54px}}~}$ $= -3 + 4 = 1$
Row 2, column 2: Hint: Row 2 of $A$ is $\begin{pmatrix} 3 & -2 \end{pmatrix}$. Column 2 of $B$ is $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1a0}}\hspace{40px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1b0}}\hspace{40px}}~} + \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1a1}}\hspace{40px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1b1}}\hspace{40px}}~}$ $3 \times 3 + -2 \times 2$
$= \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1p0}}\hspace{54px}}~} + \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1p1}}\hspace{54px}}~} = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1R}}\hspace{54px}}~}$ $= 9 + -4 = 5$
Therefore,
$AB = \begin{pmatrix} \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c0}}\hspace{54px}}~} & \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c1}}\hspace{54px}}~} \\ \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c0}}\hspace{54px}}~} & \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c1}}\hspace{54px}}~} \end{pmatrix}$ $AB = \begin{pmatrix} 5 & -7 \\ 1 & 5 \end{pmatrix}$