Formulaic Maths Problems
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Let $A = \begin{pmatrix} -7 & 0 & -6 \\ -6 & -8 & 7 \\ 0 & -12 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} 12 & -11 \\ 4 & -11 \\ -7 & -6 \end{pmatrix}$. Find $AB$.
Row 1, column 1:
Hint: Row 1 of $A$ is $\begin{pmatrix} -7 & 0 & -6 \end{pmatrix}$. Column 1 of $B$ is $\begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a0}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b0}}\hspace{40px}}~} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a1}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b1}}\hspace{40px}}~} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0a2}}\hspace{40px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0b2}}\hspace{40px}}~}$
$[-7] \times [12] + [0] \times [4] + [-6] \times [-7]$
$= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p0}}\hspace{54px}}~} + \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p1}}\hspace{54px}}~} + \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0p2}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c0R}}\hspace{54px}}~}$
$= [-84] + [0] + [42] = [-42]$
Row 1, column 2:
Hint: Row 1 of $A$ is $\begin{pmatrix} -7 & 0 & -6 \end{pmatrix}$. Column 2 of $B$ is $\begin{pmatrix} -11 \\ -11 \\ -6 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1a0}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1b0}}\hspace{40px}}~} + \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1a1}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1b1}}\hspace{40px}}~} + \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1a2}}\hspace{40px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1b2}}\hspace{40px}}~}$
$[-7] \times [-11] + [0] \times [-11] + [-6] \times [-6]$
$= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1p0}}\hspace{54px}}~} + \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1p1}}\hspace{54px}}~} + \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1p2}}\hspace{54px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r0c1R}}\hspace{54px}}~}$
$= [77] + [0] + [36] = [113]$
Row 2, column 1:
Hint: Row 2 of $A$ is $\begin{pmatrix} -6 & -8 & 7 \end{pmatrix}$. Column 1 of $B$ is $\begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a0}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b0}}\hspace{40px}}~} + \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a1}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b1}}\hspace{40px}}~} + \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0a2}}\hspace{40px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0b2}}\hspace{40px}}~}$
$[-6] \times [12] + [-8] \times [4] + [7] \times [-7]$
$= \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p0}}\hspace{54px}}~} + \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p1}}\hspace{54px}}~} + \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0p2}}\hspace{54px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c0R}}\hspace{54px}}~}$
$= [-72] + [-32] + [-49] = [-153]$
Row 2, column 2:
Hint: Row 2 of $A$ is $\begin{pmatrix} -6 & -8 & 7 \end{pmatrix}$. Column 2 of $B$ is $\begin{pmatrix} -11 \\ -11 \\ -6 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1a0}}\hspace{40px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1b0}}\hspace{40px}}~} + \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1a1}}\hspace{40px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1b1}}\hspace{40px}}~} + \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1a2}}\hspace{40px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1b2}}\hspace{40px}}~}$
$[-6] \times [-11] + [-8] \times [-11] + [7] \times [-6]$
$= \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1p0}}\hspace{54px}}~} + \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1p1}}\hspace{54px}}~} + \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1p2}}\hspace{54px}}~} = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r1c1R}}\hspace{54px}}~}$
$= [66] + [88] + [-42] = [112]$
Row 3, column 1:
Hint: Row 3 of $A$ is $\begin{pmatrix} 0 & -12 & -1 \end{pmatrix}$. Column 1 of $B$ is $\begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0a0}}\hspace{40px}}~} \times \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0b0}}\hspace{40px}}~} + \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0a1}}\hspace{40px}}~} \times \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0b1}}\hspace{40px}}~} + \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0a2}}\hspace{40px}}~} \times \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0b2}}\hspace{40px}}~}$
$[0] \times [12] + [-12] \times [4] + [-1] \times [-7]$
$= \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0p0}}\hspace{54px}}~} + \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0p1}}\hspace{54px}}~} + \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0p2}}\hspace{54px}}~} = \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c0R}}\hspace{54px}}~}$
$= [0] + [-48] + [7] = [-41]$
Row 3, column 2:
Hint: Row 3 of $A$ is $\begin{pmatrix} 0 & -12 & -1 \end{pmatrix}$. Column 2 of $B$ is $\begin{pmatrix} -11 \\ -11 \\ -6 \end{pmatrix}$. Multiply in pairs and add the results.
$\class{inputBox step11}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1a0}}\hspace{40px}}~} \times \class{inputBox step11}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1b0}}\hspace{40px}}~} + \class{inputBox step11}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1a1}}\hspace{40px}}~} \times \class{inputBox step11}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1b1}}\hspace{40px}}~} + \class{inputBox step11}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1a2}}\hspace{40px}}~} \times \class{inputBox step11}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1b2}}\hspace{40px}}~}$
$[0] \times [-11] + [-12] \times [-11] + [-1] \times [-6]$
$= \class{inputBox step12}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1p0}}\hspace{54px}}~} + \class{inputBox step12}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1p1}}\hspace{54px}}~} + \class{inputBox step12}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1p2}}\hspace{54px}}~} = \class{inputBox step12}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{r2c1R}}\hspace{54px}}~}$
$= [0] + [132] + [6] = [138]$
Therefore,
$AB = \begin{pmatrix} \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c0}}\hspace{54px}}~} & \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr0c1}}\hspace{54px}}~} \\ \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c0}}\hspace{54px}}~} & \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr1c1}}\hspace{54px}}~} \\ \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr2c0}}\hspace{54px}}~} & \class{inputBox step13}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Fr2c1}}\hspace{54px}}~} \end{pmatrix}$
$AB = \begin{pmatrix} [-42] & [113] \\ [-153] & [112] \\ [-41] & [138] \end{pmatrix}$