Formulaic Maths Problems

Find the determinant of the matrix $A = \begin{pmatrix} -1 & 4 & 2 \\ -5 & -4 & -10 \\ 5 & -7 & -6 \end{pmatrix}$ 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.

$\operatorname{det} A = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Aa}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab22}}\hspace{35px}}~} \end{pmatrix} - \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ac}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad22}}\hspace{35px}}~} \end{pmatrix} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ae}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af22}}\hspace{35px}}~} \end{pmatrix}$ $\operatorname{det} A = [-1] \operatorname{det} \begin{pmatrix} [-4] & [-10] \\ [-7] & [-6] \end{pmatrix} - [4] \operatorname{det} \begin{pmatrix} [-5] & [-10] \\ [5] & [-6] \end{pmatrix} + [2] \operatorname{det} \begin{pmatrix} [-5] & [-4] \\ [5] & [-7] \end{pmatrix}$

Calculate the first minor. Hint: The determinant of a $2{\times}2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad - bc$.

$\operatorname{det} \begin{pmatrix} -4 & -10 \\ -7 & -6 \end{pmatrix}= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B1}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B2}}\hspace{54px}}~} - \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B3}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} -4 & -10 \\ -7 & -6 \end{pmatrix}= [-4] \times [-6] - [-10] \times [-7]$

$= \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C1}}\hspace{54px}}~} - \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C2}}\hspace{54px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{54px}}~}$ $= [24] - [70] = [-46]$

Calculate the second minor.

$\operatorname{det} \begin{pmatrix} -5 & -10 \\ 5 & -6 \end{pmatrix}= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D1}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D2}}\hspace{54px}}~} - \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D3}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} -5 & -10 \\ 5 & -6 \end{pmatrix}= [-5] \times [-6] - [-10] \times [5]$

$= \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E1}}\hspace{54px}}~} - \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E2}}\hspace{54px}}~} = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{ER}}\hspace{54px}}~}$ $= [30] - [-50] = [80]$

Calculate the third minor.

$\operatorname{det} \begin{pmatrix} -5 & -4 \\ 5 & -7 \end{pmatrix}= \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F1}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F2}}\hspace{54px}}~} - \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F3}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} -5 & -4 \\ 5 & -7 \end{pmatrix}= [-5] \times [-7] - [-4] \times [5]$

$= \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{G1}}\hspace{54px}}~} - \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{G2}}\hspace{54px}}~} = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{GR}}\hspace{54px}}~}$ $= [35] - [-20] = [55]$

Therefore, Hint: Substitute the minors into the first step.

$\operatorname{det} A = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H1a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H1b}}\hspace{54px}}~} - \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H2a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H2b}}\hspace{54px}}~} + \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H3a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H3b}}\hspace{54px}}~}$ $\operatorname{det} A = [-1] \times [-46] - [4] \times [80] + [2] \times [55]$

$= \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J1}}\hspace{54px}}~} - \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J2}}\hspace{54px}}~} + \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J3}}\hspace{54px}}~} = \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{JR}}\hspace{54px}}~}$ $= [46] - [320] + [110] = [-164]$