Formulaic Maths Problems - get another problem: level 1, level 2, level 3 - link to this problem
Loading, please wait...
Find the determinant of the matrix $A = \begin{pmatrix} 10 & 1 & -12 \\ 8 & -13 & -4 \\ -8 & 15 & -16 \end{pmatrix}$ using Sarrus' rule.
Copy the columns. Hint: Copy the first two columns of $A$.
$\begin{pmatrix} 10 & 1 & -12 & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A2}}\hspace{35px}}~} \\ 8 & -13 & -4 & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A3}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A4}}\hspace{35px}}~} \\ -8 & 15 & -16 & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A5}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A6}}\hspace{35px}}~} \end{pmatrix}$ $\begin{pmatrix} 10 & 1 & -12 & [10] & [1] \\ 8 & -13 & -4 & [8] & [-13] \\ -8 & 15 & -16 & [-8] & [15] \end{pmatrix}$
Multiply the $\searrow$ diagonals.
$\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ba1}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ba2}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ba3}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BaR}}\hspace{54px}}~}$ $[10] \times [-13] \times [-16] = [2080]$
$\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bb1}}\hspace{54px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bb2}}\hspace{54px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bb3}}\hspace{54px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BbR}}\hspace{54px}}~}$ $[1] \times [-4] \times [-8] = [32]$
$\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bc1}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bc2}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Bc3}}\hspace{54px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BcR}}\hspace{54px}}~}$ $[-12] \times [8] \times [15] = [-1440]$
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 ${[672]}$
Multiply the $\swarrow$ diagonals.
$\class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ca1}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ca2}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ca3}}\hspace{54px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CaR}}\hspace{54px}}~}$ $[-12] \times [-13] \times [-8] = [-1248]$
$\class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cb1}}\hspace{54px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cb2}}\hspace{54px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cb3}}\hspace{54px}}~} = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CbR}}\hspace{54px}}~}$ $[10] \times [-4] \times [15] = [-600]$
$\class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cc1}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cc2}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Cc3}}\hspace{54px}}~} = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CcR}}\hspace{54px}}~}$ $[1] \times [8] \times [-16] = [-128]$
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 ${[-1976]}$
Therefore, Hint: The determinant is the total of the $\searrow$ diagonals, minus the total of the $\swarrow$ diagonals.
$\operatorname{det} A = \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D1}}\hspace{54px}}~} - \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D2}}\hspace{54px}}~} = \class{inputBox step10}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{DR}}\hspace{54px}}~}$ $\operatorname{det} A = [672] - [-1976] = [2648]$