question_answer

Find the adjoint of the matrix and hence show that A (adj A) \[=\,\,|A|{{I}_{3}}.\]

Answer:

We have, Now, \[=-\,1\,(1-4)-(-\,2)\,\,(2+4)+(-\,2)\,(-\,4-2)\] \[=3+12+12=27\] Cofactors of A are \[{{A}_{11}}={{(-\,1)}^{1\,\,+\,\,1}}\left| \begin{matrix}    1 & -\,2  \\    -\,2 & 1  \\ \end{matrix} \right|=1\,(1-4)=-\,3\] \[{{A}_{12}}={{(-\,1)}^{1\,\,+\,\,2}}\left| \begin{matrix}    2 & -\,2  \\    2 & 1  \\ \end{matrix} \right|=(-1)\,\,(2+4)=-\,6\] \[{{A}_{13}}={{(-\,1)}^{1\,\,+\,\,3}}\left| \begin{matrix}    2 & 1  \\    2 & -\,2  \\ \end{matrix} \right|=1\,(-\,4-2)=-\,6\] \[{{A}_{21}}={{(-\,1)}^{2\,\,+\,\,1}}\left| \begin{matrix}    -\,2 & -\,2  \\    -\,2 & 1  \\ \end{matrix} \right|=(-\,1)\,\,(-\,2-4)=6\] \[{{A}_{22}}={{(-\,1)}^{2\,\,+\,\,2}}\left| \begin{matrix}    -\,1 & -\,2  \\    2 & 1  \\ \end{matrix} \right|=1\,\,(-\,1+4)=3\] \[{{A}_{23}}={{(-\,1)}^{2\,\,+\,\,3}}\left| \begin{matrix}    -\,1 & -\,2  \\    2 & -\,2  \\ \end{matrix} \right|=(-\,1)\,\,(2+4)=-\,6\] \[{{A}_{31}}={{(-\,1)}^{3\,\,+\,\,1}}\left| \begin{matrix}    -\,2 & -\,2  \\    1 & -\,2  \\ \end{matrix} \right|=1\,(4+2)=6\] \[{{A}_{32}}={{(-\,1)}^{3\,\,+\,\,2}}\left| \begin{matrix}    -\,1 & -\,2  \\    2 & -\,2  \\ \end{matrix} \right|=(-\,1)\,\,(2+4)=-\,6\] \[{{A}_{33}}={{(-\,1)}^{3\,\,+\,\,3}}\left| \begin{matrix}    -\,1 & -\,2  \\    2 & 1  \\ \end{matrix} \right|=1\,(-\,1+4)=3\] \[\therefore Now,      And Hence, \[A(adj\,A)=|A|{{l}_{3}}\]                   Hence proved.