Question

Solve it.

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Answer 1

Answer:

10

Step-by-step explanation:

1+2-3+4+4+2

=10

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Answer 2

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

Given matrix is

$\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}1&2& - 3\\ - 3&2&4\\ 6&8&4\end{array}\right]\end{gathered}$

Now,

$\red{\rm :\longmapsto\:A_{11} = {( - 1)}^{1 + 1}\begin{array}{|cc|}\sf 2 &\sf 4 \\ \sf 8 &\sf 4 \\\end{array} = 8 - 32 = - 24 \: }$

$\green{\rm :\longmapsto\:A_{12} = {( - 1)}^{1 + 2}\begin{array}{|cc|}\sf - 3 &\sf 4 \\ \sf 6 &\sf 4 \\\end{array} = - ( - 12 - 24) = 36 \: }$

$\blue{\rm :\longmapsto\:A_{13} = {( - 1)}^{1 + 3}\begin{array}{|cc|}\sf - 3 &\sf 2 \\ \sf 6 &\sf 8 \\\end{array} = - 24 - 12 = -36 \: }$

$\purple{\rm :\longmapsto\:A_{22} = {( - 1)}^{2 + 2}\begin{array}{|cc|}\sf 1 &\sf - 3 \\ \sf 6 &\sf 4 \\\end{array} = 4 + 18 =22\: }$

$\pink{\rm :\longmapsto\:A_{33} = {( - 1)}^{3 + 3}\begin{array}{|cc|}\sf 1 &\sf 2 \\ \sf - 3 &\sf 2 \\\end{array} = 2 + 6 =8\: }$

$\pink{\rm :\longmapsto\:A_{23} = {( - 1)}^{2+ 3}\begin{array}{|cc|}\sf 1 &\sf 2 \\ \sf 6 &\sf 8 \\\end{array} = - (8 - 12) =4\: }$

Now, Consider

$\rm :\longmapsto\:A_{11} + A_{12} + A_{13} + A_{22} + A_{33} + A_{23}$

$\rm \:  =  \: - 24 + 36 - 36 + 22 + 8 + 4$

$\rm \:  =  \: 10$

Hence,

$\boxed{ \tt{ \: \:A_{11} + A_{12} + A_{13} + A_{22} + A_{33} + A_{23} = 10 \: \: }}$

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Additional Information :-

$\boxed{ \tt{ \: a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0}}$

$\boxed{ \tt{ \: a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = |A| }}$

$\boxed{ \tt{ \: a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = |A| }}$

$\boxed{ \tt{ \: a_{31}A_{31} + a_{32}A_{32} + a_{33}A_{33} = |A| }}$

$\boxed{ \tt{ \: A_{ij} \: = \: {( - 1)}^{i \: + \: j} \: m_{ij} \: }}$