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Find the determinant of the matrix $A = \begin{pmatrix} 9 & 7 & -12 \\ -6 & 6 & 5 \\ -1 & 1 & 1 \end{pmatrix}$ using Sarrus' rule.
Copy the columns. Hint: Copy the first two columns of $A$.
$\begin{pmatrix} 9 & 7 & -12 & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A2}}\hspace{35px}}~} \\ -6 & 6 & 5 & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A3}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A4}}\hspace{35px}}~} \\ -1 & 1 & 1 & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A5}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A6}}\hspace{35px}}~} \end{pmatrix}$ $\begin{pmatrix} 9 & 7 & -12 & 9 & 7 \\ -6 & 6 & 5 & -6 & 6 \\ -1 & 1 & 1 & -1 & 1 \end{pmatrix}$
Multiply the $\searrow$ diagonals.
$\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ba1}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ba2}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ba3}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BaR}}\hspace{54px}}~}$ $9 \times 6 \times 1 = 54$
$\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bb1}}\hspace{54px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bb2}}\hspace{54px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bb3}}\hspace{54px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BbR}}\hspace{54px}}~}$ $7 \times 5 \times -1 = -35$
$\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bc1}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bc2}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bc3}}\hspace{54px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BcR}}\hspace{54px}}~}$ $-12 \times -6 \times 1 = 72$
Hint: Add together the results from the $\searrow$ diagonals.
So the total is ${\class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BT}}\hspace{54px}}~}}$ So the total is ${91}$
Multiply the $\swarrow$ diagonals.
$\class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ca1}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ca2}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ca3}}\hspace{54px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CaR}}\hspace{54px}}~}$ $-12 \times 6 \times -1 = 72$
$\class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cb1}}\hspace{54px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cb2}}\hspace{54px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cb3}}\hspace{54px}}~} = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CbR}}\hspace{54px}}~}$ $9 \times 5 \times 1 = 45$
$\class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cc1}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cc2}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cc3}}\hspace{54px}}~} = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CcR}}\hspace{54px}}~}$ $7 \times -6 \times 1 = -42$
Hint: Add together the results from the $\swarrow$ diagonals.
So the total is ${\class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CT}}\hspace{54px}}~}}$ So the total is ${75}$
Therefore, Hint: The determinant is the total of the $\searrow$ diagonals, minus the total of the $\swarrow$ diagonals.
$\operatorname{det} A = \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D1}}\hspace{54px}}~} - \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D2}}\hspace{54px}}~} = \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{DR}}\hspace{54px}}~}$ $\operatorname{det} A = 91 - 75 = 16$