Question

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

Answer 1

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

Given matrix is

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

Let assume that

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

Now, Lets evaluate Cofactors

$\red{\rm :\longmapsto\:a_{11} = {( - 1)}^{1 + 1}\begin{array}{|cc|}\sf 1 &\sf - 1 \\ \sf - 3 &\sf 2 \\\end{array} = 2 - 3 = - 1}$

$\red{\rm :\longmapsto\:a_{12} = {( - 1)}^{1 + 2}\begin{array}{|cc|}\sf 2 &\sf - 1 \\ \sf 4 &\sf 2 \\\end{array} = - (4 + 4) = - 8}$

$\red{\rm :\longmapsto\:a_{13} = {( - 1)}^{1 + 3}\begin{array}{|cc|}\sf 2 &\sf 1 \\ \sf 4 &\sf - 3 \\\end{array} = -6 - 4 = - 10}$

$\green{\rm :\longmapsto\:a_{21} = {( - 1)}^{2 + 1}\begin{array}{|cc|}\sf - 2 &\sf 3 \\ \sf - 3 &\sf 2 \\\end{array} = - ( - 4 + 9) = - 5}$

$\green{\rm :\longmapsto\:a_{22} = {( - 1)}^{2 + 2}\begin{array}{|cc|}\sf 3 &\sf 3 \\ \sf 4 &\sf 2 \\\end{array} = 6 - 12 = - 6}$

$\green{\rm :\longmapsto\:a_{23} = {( - 1)}^{2 + 3}\begin{array}{|cc|}\sf 3 &\sf - 2 \\ \sf 4 &\sf - 3 \\\end{array} = - ( - 9 + 8) =1}$

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

$\blue{\rm :\longmapsto\:a_{32} = {( - 1)}^{3 + 2}\begin{array}{|cc|}\sf 3 &\sf 3 \\ \sf 2 &\sf - 1 \\\end{array} = - ( - 3 - 6)= 9}$

$\blue{\rm :\longmapsto\:a_{33} = {( - 1)}^{3 + 3}\begin{array}{|cc|}\sf 3 &\sf - 2 \\ \sf 2 &\sf 1 \\\end{array} = 3 + 4= 7}$

So,

$\red{\rm :\longmapsto\:adj(A) = \left[ \begin{array} { l l l } { - 1 } & { - 8 } & { - 10 } \\ { - 5 } & { - 6 } & { 1 } \\ { - 1 } & { 9 } & { 7 } \end{array} \right]'}$

$\red{\rm :\longmapsto\:adjA = \left[ \begin{array} { l l l } { - 1 } & { - 5 } & { - 1 } \\ { - 8 } & { - 6 } & {9 } \\ { - 10} & { 1 } & { 7 } \end{array} \right]}$

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Explore More :-

$\boxed{\tt{ A \: adj(A) \: = \: adj(A) \: A \: = \: |A| I\: }}$

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

$\boxed{\tt{ |k \: A| = {k}^{n} |A| \: }}$

Answer 2

Answer:

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