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Find the determinant of the matrix $A = \begin{pmatrix} 9 & -8 & -1 \\ -4 & 12 & 5 \\ -6 & 10 & -12 \end{pmatrix}$ by expansion of cofactors.
Expand. Hint: Multiply each element from the first row by its minor. The minor is the determinant of the matrix found by deleting that row and column.
$\operatorname{det} A = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Aa}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab22}}\hspace{35px}}~} \end{pmatrix} - \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ac}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad22}}\hspace{35px}}~} \end{pmatrix} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ae}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af22}}\hspace{35px}}~} \end{pmatrix}$ $\operatorname{det} A = [9] \operatorname{det} \begin{pmatrix} [12] & [5] \\ [10] & [-12] \end{pmatrix} - [-8] \operatorname{det} \begin{pmatrix} [-4] & [5] \\ [-6] & [-12] \end{pmatrix} + [-1] \operatorname{det} \begin{pmatrix} [-4] & [12] \\ [-6] & [10] \end{pmatrix}$
Calculate the first minor. Hint: The determinant of a $2{\times}2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad - bc$.
$\operatorname{det} \begin{pmatrix} 12 & 5 \\ 10 & -12 \end{pmatrix}= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B1}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B2}}\hspace{54px}}~} - \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B3}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} 12 & 5 \\ 10 & -12 \end{pmatrix}= [12] \times [-12] - [5] \times [10]$
$= \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C1}}\hspace{54px}}~} - \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C2}}\hspace{54px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{54px}}~}$ $= [-144] - [50] = [-194]$
Calculate the second minor.
$\operatorname{det} \begin{pmatrix} -4 & 5 \\ -6 & -12 \end{pmatrix}= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D1}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D2}}\hspace{54px}}~} - \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D3}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} -4 & 5 \\ -6 & -12 \end{pmatrix}= [-4] \times [-12] - [5] \times [-6]$
$= \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E1}}\hspace{54px}}~} - \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E2}}\hspace{54px}}~} = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{ER}}\hspace{54px}}~}$ $= [48] - [-30] = [78]$
Calculate the third minor.
$\operatorname{det} \begin{pmatrix} -4 & 12 \\ -6 & 10 \end{pmatrix}= \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F1}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F2}}\hspace{54px}}~} - \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F3}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} -4 & 12 \\ -6 & 10 \end{pmatrix}= [-4] \times [10] - [12] \times [-6]$
$= \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{G1}}\hspace{54px}}~} - \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{G2}}\hspace{54px}}~} = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{GR}}\hspace{54px}}~}$ $= [-40] - [-72] = [32]$
Therefore, Hint: Substitute the minors into the first step.
$\operatorname{det} A = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H1a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H1b}}\hspace{54px}}~} - \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H2a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H2b}}\hspace{54px}}~} + \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H3a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H3b}}\hspace{54px}}~}$ $\operatorname{det} A = [9] \times [-194] - [-8] \times [78] + [-1] \times [32]$
$= \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J1}}\hspace{54px}}~} - \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J2}}\hspace{54px}}~} + \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J3}}\hspace{54px}}~} = \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{JR}}\hspace{54px}}~}$ $= [-1746] - [-624] + [-32] = [-1154]$