Formulaic Maths Problems

Find the determinant of the matrix $A = \begin{pmatrix} 12 & -7 & -11 \\ 2 & 9 & 8 \\ 12 & -9 & 0 \end{pmatrix}$ 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.

$\operatorname{det} A = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Aa}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ab22}}\hspace{35px}}~} \end{pmatrix} - \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ac}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ad22}}\hspace{35px}}~} \end{pmatrix} + \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ae}}\hspace{35px}}~} \operatorname{det} \begin{pmatrix} \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af11}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af12}}\hspace{35px}}~} \\ \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af21}}\hspace{35px}}~} & \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Af22}}\hspace{35px}}~} \end{pmatrix}$ $\operatorname{det} A = [12] \operatorname{det} \begin{pmatrix} [9] & [8] \\ [-9] & [0] \end{pmatrix} - [-7] \operatorname{det} \begin{pmatrix} [2] & [8] \\ [12] & [0] \end{pmatrix} + [-11] \operatorname{det} \begin{pmatrix} [2] & [9] \\ [12] & [-9] \end{pmatrix}$

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

$\operatorname{det} \begin{pmatrix} 9 & 8 \\ -9 & 0 \end{pmatrix}= \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B1}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B2}}\hspace{54px}}~} - \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B3}}\hspace{54px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} 9 & 8 \\ -9 & 0 \end{pmatrix}= [9] \times [0] - [8] \times [-9]$

$= \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C1}}\hspace{54px}}~} - \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C2}}\hspace{54px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{54px}}~}$ $= [0] - [-72] = [72]$

Calculate the second minor.

$\operatorname{det} \begin{pmatrix} 2 & 8 \\ 12 & 0 \end{pmatrix}= \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D1}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D2}}\hspace{54px}}~} - \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D3}}\hspace{54px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} 2 & 8 \\ 12 & 0 \end{pmatrix}= [2] \times [0] - [8] \times [12]$

$= \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E1}}\hspace{54px}}~} - \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E2}}\hspace{54px}}~} = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{ER}}\hspace{54px}}~}$ $= [0] - [96] = [-96]$

Calculate the third minor.

$\operatorname{det} \begin{pmatrix} 2 & 9 \\ 12 & -9 \end{pmatrix}= \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F1}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F2}}\hspace{54px}}~} - \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F3}}\hspace{54px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F4}}\hspace{54px}}~}$ $\operatorname{det} \begin{pmatrix} 2 & 9 \\ 12 & -9 \end{pmatrix}= [2] \times [-9] - [9] \times [12]$

$= \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{G1}}\hspace{54px}}~} - \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{G2}}\hspace{54px}}~} = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{GR}}\hspace{54px}}~}$ $= [-18] - [108] = [-126]$

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

$\operatorname{det} A = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H1a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H1b}}\hspace{54px}}~} - \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H2a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H2b}}\hspace{54px}}~} + \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H3a}}\hspace{54px}}~} \times \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{H3b}}\hspace{54px}}~}$ $\operatorname{det} A = [12] \times [72] - [-7] \times [-96] + [-11] \times [-126]$

$= \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J1}}\hspace{54px}}~} - \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J2}}\hspace{54px}}~} + \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{J3}}\hspace{54px}}~} = \class{inputBox step9}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{JR}}\hspace{54px}}~}$ $= [864] - [672] + [1386] = [1578]$