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Find the determinant of the matrix $A = \begin{pmatrix} -4 & -5 & 3 \\ -6 & 6 & -1 \\ -5 & -3 & -5 \end{pmatrix}$ 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.
$\operatorname{det} A = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Aa}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab22}}\hspace{35px}}~} \end{pmatrix} - \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ac}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad22}}\hspace{35px}}~} \end{pmatrix} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ae}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af22}}\hspace{35px}}~} \end{pmatrix}$ $\operatorname{det} A = [-4] \operatorname{det} \begin{pmatrix} [6] & [-1] \\ [-3] & [-5] \end{pmatrix} - [-5] \operatorname{det} \begin{pmatrix} [-6] & [-1] \\ [-5] & [-5] \end{pmatrix} + [3] \operatorname{det} \begin{pmatrix} [-6] & [6] \\ [-5] & [-3] \end{pmatrix}$
Calculate the first minor. Hint: The determinant of a $2{\times}2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad - bc$.
$\operatorname{det} \begin{pmatrix} 6 & -1 \\ -3 & -5 \end{pmatrix}= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B1}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B2}}\hspace{54px}}~} - \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B3}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} 6 & -1 \\ -3 & -5 \end{pmatrix}= [6] \times [-5] - [-1] \times [-3]$
$= \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C1}}\hspace{54px}}~} - \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C2}}\hspace{54px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{54px}}~}$ $= [-30] - [3] = [-33]$
Calculate the second minor.
$\operatorname{det} \begin{pmatrix} -6 & -1 \\ -5 & -5 \end{pmatrix}= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D1}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D2}}\hspace{54px}}~} - \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D3}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} -6 & -1 \\ -5 & -5 \end{pmatrix}= [-6] \times [-5] - [-1] \times [-5]$
$= \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E1}}\hspace{54px}}~} - \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E2}}\hspace{54px}}~} = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{ER}}\hspace{54px}}~}$ $= [30] - [5] = [25]$
Calculate the third minor.
$\operatorname{det} \begin{pmatrix} -6 & 6 \\ -5 & -3 \end{pmatrix}= \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F1}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F2}}\hspace{54px}}~} - \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F3}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} -6 & 6 \\ -5 & -3 \end{pmatrix}= [-6] \times [-3] - [6] \times [-5]$
$= \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{G1}}\hspace{54px}}~} - \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{G2}}\hspace{54px}}~} = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{GR}}\hspace{54px}}~}$ $= [18] - [-30] = [48]$
Therefore, Hint: Substitute the minors into the first step.
$\operatorname{det} A = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H1a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H1b}}\hspace{54px}}~} - \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H2a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H2b}}\hspace{54px}}~} + \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H3a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H3b}}\hspace{54px}}~}$ $\operatorname{det} A = [-4] \times [-33] - [-5] \times [25] + [3] \times [48]$
$= \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J1}}\hspace{54px}}~} - \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J2}}\hspace{54px}}~} + \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J3}}\hspace{54px}}~} = \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{JR}}\hspace{54px}}~}$ $= [132] - [-125] + [144] = [401]$