Question

find the adjoint of the matrix: \
$\mathtt\purple{\left[ \begin{array} { l l l } { 2 } & { - 1 } & { 3 } \\ { 4 } & { 2 } & { 5 } \\ { 0 } & { 4 } & { 1 } \end{array} \right]}$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given matrix is

$\rm :\longmapsto\:{\left[ \begin{array} { l l l } { 2 } & { - 1 } & { 3 } \\ { 4 } & { 2 } & { 5 } \\ { 0 } & { 4 } & { 1 } \end{array} \right]}$

Let assume that

$\rm :\longmapsto\:A = {\left[ \begin{array} { l l l } { 2 } & { - 1 } & { 3 } \\ { 4 } & { 2 } & { 5 } \\ { 0 } & { 4 } & { 1 } \end{array} \right]}$

Let evaluate the cofactors of matrix A

$\red{\rm :\longmapsto\:A_{11} = {( - 1)}^{1 + 1}\begin{array}{|cc|}\sf 2 &\sf 5 \\ \sf 4 &\sf 1 \\\end{array} = 2 - 20 = - 18}$

$\red{\rm :\longmapsto\:A_{12} = {( - 1)}^{1 + 2}\begin{array}{|cc|}\sf 4 &\sf 5 \\ \sf 0 &\sf 1 \\\end{array} = - (4 - 0) = - 4}$

$\red{\rm :\longmapsto\:A_{13} = {( - 1)}^{1 + 3}\begin{array}{|cc|}\sf 4 &\sf 2 \\ \sf 0 &\sf 4 \\\end{array} = 16 - 0 = 16}$

$\green{\rm :\longmapsto\:A_{21} = {( - 1)}^{2 + 1}\begin{array}{|cc|}\sf - 1 &\sf 3 \\ \sf 4 &\sf 1 \\\end{array} = - ( - 1 - 12) = 13 }$

$\green{\rm :\longmapsto\:A_{22} = {( - 1)}^{2 + 2}\begin{array}{|cc|}\sf 2 &\sf 3 \\ \sf 0 &\sf 1 \\\end{array} = 2 - 0 = 2}$

$\green{\rm :\longmapsto\:A_{23} = {( - 1)}^{2 + 3}\begin{array}{|cc|}\sf 2 &\sf - 1 \\ \sf 0 &\sf 4 \\\end{array} = - (8 - 0) = - 8}$

$\blue{\rm :\longmapsto\:A_{31} = {( - 1)}^{3 + 1}\begin{array}{|cc|}\sf - 1 &\sf 3 \\ \sf 2 &\sf 5 \\\end{array} = - 5 - 6 = - 11 }$

$\blue{\rm :\longmapsto\:A_{32} = {( - 1)}^{3 + 2}\begin{array}{|cc|}\sf 2 &\sf 3 \\ \sf 4 &\sf 5 \\\end{array} = - (10 - 12) = 2}$

$\blue{\rm :\longmapsto\:A_{33} = {( - 1)}^{3 + 3}\begin{array}{|cc|}\sf 2 &\sf - 1 \\ \sf 4 &\sf 2 \\\end{array} = 4 + 4 = 8}$

So,

$\bf\implies \:adjA = {\left[ \begin{array} { l l l } { - 18 } & { 13 } & { - 11 } \\ { - 4 } & { 2 } & { 2 } \\ { 16 } & { - 8 } & { 8 } \end{array} \right]}$

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Additional Information

Let us consider a matrix A of order n such that

$\red{\rm :\longmapsto\:\boxed{\tt{ |adjA| = { |A| }^{n - 1}}}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ |AadjA| = { |A| }^{n}}}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ |kA| = {k}^{n} |A| \: }}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ |AB| \: = \: |A| \: |B| \: }}}$

Answer 2

Answer:

Write shortcut key to save and open an existing database in .accdb. Refer to the above attachment too dear