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# If ${\text{A = }}\left( {\begin{array}{*{20}{c}} {\text{2}}&{\text{0}}&{{\text{ - 1}}} \\ {\text{5}}&{\text{1}}&{\text{0}} \\ {\text{0}}&{\text{1}}&{\text{3}} \end{array}} \right)$ and then find the value of ${{\text{A}}^{{\text{ - 1}}}} =$ A) ${{\text{A}}^{\text{2}}}{\text{ - 6A + 11I}}$ B) ${{\text{A}}^{\text{2}}}{\text{ + 6A - 11I}}$ C) ${{\text{A}}^{\text{2}}}{\text{ - 6A - 11I}}$ D) ${{\text{A}}^{\text{2}}}{\text{ + 6A + 11I}}$

Last updated date: 22nd Jun 2024
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Hint: In this question we have found the value of inverse of the given matrix. For that we solve this by using matrices method. First we are going to find the modulus of a given matrix and adjacent of a given matrix. Next, we have to check which option is the right one. So we are going to solve one by one. Commonly first find the ${{\text{A}}^2}$ next we know that identity matrices.

Formula used: ${{\text{A}}^{{\text{ - 1}}}}{\text{ = }}\dfrac{{{\text{adj(A)}}}}{{{\text{|A|}}}}$
${\text{adj}}\left( {\text{A}} \right) = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right)$

It is given that, ${\text{A = }}\left( {\begin{array}{*{20}{c}} {\text{2}}&{\text{0}}&{{\text{ - 1}}} \\ {\text{5}}&{\text{1}}&{\text{0}} \\ {\text{0}}&{\text{1}}&{\text{3}} \end{array}} \right)$
Now to find the determinant of the matrix ${\text{A}}$,
${\text{|A| = 2}}\left[ {\left( {{\text{1}} \times {\text{3}}} \right) - \left( {1 \times 0} \right)} \right] - {\text{0}}\left[ {\left( {5 \times 3} \right) - \left( {0 \times 0} \right)} \right]{\text{ + }}\left( { - 1} \right)\left[ {\left( {5 \times 1} \right) - \left( {0 \times 1} \right)} \right]$
Multiplying the terms we get,
${\text{|A| = 2}}\left[ {\left( {\text{3}} \right) - \left( 0 \right)} \right] - {\text{0}}\left[ {\left( {15} \right) - \left( 0 \right)} \right] - 1\left[ {\left( 5 \right) - \left( 0 \right)} \right]$
Simplifying we get,
${\text{|A| = 2(3) - 0 - 1(5)}}$
Multiplying the terms we get,
${\text{|A| = 6}} - 5$
${\text{|A| = 1 }} \ne {\text{ 0}}$(If ${\text{|A| = 0}}$there is no solution)
Now consider the general form of adjacent matrix of a given matrix,
${\text{adj}}\left( {\text{A}} \right) = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right)$
Now we have to find cofactor of each element,
First ${a_{11}}$has been chosen as cancelling first row and first column and proceeding for remaining values like this. ${a_{11}} = \left( {\begin{array}{*{20}{c}} 1&0 \\ 1&3 \end{array}} \right) = 3$
For ${a_{12}}$has been chosen as cancelling first row and second column and proceeding for remaining values like this. ${a_{12}} = - \left( {\begin{array}{*{20}{c}} 0&{ - 1} \\ 1&3 \end{array}} \right) = - 1$
For ${a_{13}}$ has been chosen as cancelling first row and third column and proceeding for remaining values like this. ${a_{13}} = \left( {\begin{array}{*{20}{c}} 0&{ - 1} \\ 1&0 \end{array}} \right) = 1$
For ${a_{21}}$ has been chosen as cancelling second row and first column and proceeding for remaining values like this. ${a_{21}} = - \left( {\begin{array}{*{20}{c}} 5&0 \\ 0&3 \end{array}} \right) = - 15$
For ${a_{22}}$has been chosen as cancelling second row and second column and proceeding for remaining values like this. ${a_{22}} = \left( {\begin{array}{*{20}{c}} 2&{ - 1} \\ 0&3 \end{array}} \right) = 6$
For ${a_{23}}$ has been chosen as cancelling second row and third column and proceeding for remaining values like this. ${a_{23}} = - \left( {\begin{array}{*{20}{c}} 2&{ - 1} \\ 5&0 \end{array}} \right) = - 5$
For ${a_{31}}$ has been chosen as cancelling third row and first column and proceeding for remaining values like this. ${a_{31}} = \left( {\begin{array}{*{20}{c}} 5&1 \\ 0&1 \end{array}} \right) = 5$
For ${a_{32}}$ has been chosen as cancelling third row and second column and proceeding for remaining values like this. ${a_{32}} = - \left( {\begin{array}{*{20}{c}} 2&0 \\ 0&1 \end{array}} \right) = - 2$
For ${a_{33}}$ has been chosen as cancelling third row and third column and proceeding for remaining values like this. ${a_{33}} = \left( {\begin{array}{*{20}{c}} 2&0 \\ 0&1 \end{array}} \right) = 2$
Hence substituting the values we get,
${\text{adj}}\left( {\text{A}} \right) = \left( {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\ { - 15}&6&{ - 5} \\ 5&{ - 2}&2 \end{array}} \right)$
Since, ${{\text{A}}^{{\text{ - 1}}}}{\text{ = }}\dfrac{{{\text{adj(A)}}}}{{{\text{|A|}}}}$ by using we get,
${{\text{A}}^{{\text{ - 1}}}} = \dfrac{{\left( {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\ { - 15}&6&{ - 5} \\ 5&{ - 2}&2 \end{array}} \right)}}{1}$
Simplifying we get,
${{\text{A}}^{{\text{ - 1}}}} = \left( {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\ { - 15}&6&{ - 5} \\ 5&{ - 2}&2 \end{array}} \right)$
Hence we found the inverse matrix of a given matrix $A$. The exact problem is, we have to choose the given option of expressions which is equivalent to the inverse matrix of a given matrix $A$. For that we have checked each expression by substituting the matrix value of $A$ and ${A^{ - 1}}$ into that expression.
For expression of option A) we have to find ${{\text{A}}^{\text{2}}}{\text{ - 6A + 11I}}$ by substituting the matrix value of $A$ and ${A^{ - 1}}$ into that expression.
${{\text{A}}^{\text{2}}}{{ = A \times A = }}\left( {\begin{array}{*{20}{c}} {\text{2}}&{\text{0}}&{{\text{ - 1}}} \\ {\text{5}}&{\text{1}}&{\text{0}} \\ {\text{0}}&{\text{1}}&{\text{3}} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} {\text{2}}&{\text{0}}&{{\text{ - 1}}} \\ {\text{5}}&{\text{1}}&{\text{0}} \\ {\text{0}}&{\text{1}}&{\text{3}} \end{array}} \right)$
First row and first column multiply and put the values in the first row. Next second row and second column multiply and put the values in second row, third row and third column multiply and put the values in third row.
${{\text{A}}^{\text{2}}} = \left( {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 5} \\ {15}&1&{ - 5} \\ 5&4&9 \end{array}} \right)$
Substituting the values of $A$ and ${A^{ - 1}}$ into that expression we get,
$I$ is an identity matrix. That is, $I = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$
${{\text{A}}^{\text{2}}}{\text{ - 6A + 11I}} = \left( {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 5} \\ {15}&1&{ - 5} \\ 5&4&9 \end{array}} \right) - 6\left( {\begin{array}{*{20}{c}} 2&0&{ - 1} \\ 5&1&0 \\ 0&1&3 \end{array}} \right) + 11\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$
Simplifying matrices we get,
$\Rightarrow \left( {\begin{array}{*{20}{c}} {4 - 12 + 11}&{ - 1 - 0 + 0}&{ - 5 + 6 + 0} \\ {15 - 30 + 0}&{1 - 6 + 11}&{ - 5 - 0 + 0} \\ {5 - 0 + 0}&{4 - 6 + 0}&{9 - 18 + 11} \end{array}} \right)$
$\Rightarrow \left( {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\ { - 15}&6&{ - 5} \\ 5&{ - 2}&2 \end{array}} \right)$
Hence we get,
A) ${{\text{A}}^{\text{2}}}{\text{ - 6A + 11I}} = \left( {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\ { - 15}&6&{ - 5} \\ 5&{ - 2}&2 \end{array}} \right)$
For expression of option B) we have to find ${{\text{A}}^{\text{2}}}{\text{ + 6A - 11I}}$by substituting the matrix value of $A$ and ${A^{ - 1}}$ into that expression. Hence,
${{\text{A}}^{\text{2}}} = \left( {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 5} \\ {15}&1&{ - 5} \\ 5&4&9 \end{array}} \right)$
$I$ is an identity matrix. That is, $I = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$
Substituting the values of $A$ and ${A^{ - 1}}$ into that expression we get,
${{\text{A}}^{\text{2}}}{\text{ + 6A - 11I}} = \left( {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 5} \\ {15}&1&{ - 5} \\ 5&4&9 \end{array}} \right) + 6\left( {\begin{array}{*{20}{c}} 2&0&{ - 1} \\ 5&1&0 \\ 0&1&3 \end{array}} \right) - 11\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$
Simplifying matrices we get,
$\Rightarrow \left( {\begin{array}{*{20}{c}} {4 + 12 - 11}&{ - 1 + 0 + 0}&{ - 5 - 6 - 0} \\ {15 + 30 + 0}&{1 + 6 - 11}&{ - 5 + 0 + 0} \\ {5 + 0 + 0}&{4 + 6 + 0}&{9 + 18 - 11} \end{array}} \right)$
$\Rightarrow \left( {\begin{array}{*{20}{c}} 5&{ - 1}&{ - 11} \\ {45}&{ - 4}&{ - 5} \\ 5&{10}&{15} \end{array}} \right)$
Hence we get,
B) ${{\text{A}}^{\text{2}}}{\text{ + 6A - 11I}} = \left( {\begin{array}{*{20}{c}} 5&{ - 1}&{ - 11} \\ {45}&{ - 4}&{ - 5} \\ 5&{10}&{15} \end{array}} \right)$
For expression of option C) we have to find ${{\text{A}}^{\text{2}}}{\text{ - 6A - 11I}}$ by substituting the matrix value of $A$ and ${A^{ - 1}}$ into that expression. Hence,
${{\text{A}}^{\text{2}}} = \left( {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 5} \\ {15}&1&{ - 5} \\ 5&4&9 \end{array}} \right)$
$I$ is an identity matrix. That is, $I = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$
Substituting the values of $A$ and ${A^{ - 1}}$ into that expression we get,
${{\text{A}}^{\text{2}}}{\text{ - 6A - 11I}} = \left( {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 5} \\ {15}&1&{ - 5} \\ 5&4&9 \end{array}} \right) - 6\left( {\begin{array}{*{20}{c}} 2&0&{ - 1} \\ 5&1&0 \\ 0&1&3 \end{array}} \right) - 11\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$
Simplifying matrices we get,
$\Rightarrow \left( {\begin{array}{*{20}{c}} {4 - 12 - 11}&{ - 1 - 0 + 0}&{ - 5 + 6 - 0} \\ {15 - 30 - 0}&{1 - 6 - 11}&{ - 5 - 0 - 0} \\ {5 - 0 - 0}&{4 - 6 - 0}&{9 - 18 - 11} \end{array}} \right)$
$\Rightarrow \left( {\begin{array}{*{20}{c}} { - 19}&{ - 1}&1 \\ { - 15}&{ - 16}&{ - 5} \\ { - 5}&{ - 2}&{ - 20} \end{array}} \right)$
Hence we get,
C) ${{\text{A}}^{\text{2}}}{\text{ - 6A - 11I}} = \left( {\begin{array}{*{20}{c}} { - 19}&{ - 1}&1 \\ { - 15}&{ - 16}&{ - 5} \\ { - 5}&{ - 2}&{ - 20} \end{array}} \right)$
For expression of option D) we have to find ${{\text{A}}^{\text{2}}}{\text{ + 6A + 11I}}$ by substituting the matrix value of $A$ and ${A^{ - 1}}$ into that expression. Hence,
${{\text{A}}^{\text{2}}} = \left( {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 5} \\ {15}&1&{ - 5} \\ 5&4&9 \end{array}} \right)$
$I$ is an identity matrix. That is, $I = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$
Substituting the values of $A$ and ${A^{ - 1}}$ into that expression we get,
${{\text{A}}^{\text{2}}}{\text{ + 6A + 11I}} = \left( {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 5} \\ {15}&1&{ - 5} \\ 5&4&9 \end{array}} \right) + 6\left( {\begin{array}{*{20}{c}} 2&0&{ - 1} \\ 5&1&0 \\ 0&1&3 \end{array}} \right) + 11\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$
Simplifying matrices we get,
$\Rightarrow \left( {\begin{array}{*{20}{c}} {4 + 12 + 11}&{ - 1 + 0 + 0}&{ - 5 + 6 + 0} \\ {15 + 30 - 0}&{1 + 6 + 11}&{ - 5 + 0 + 0} \\ {5 + 0 + 0}&{4 + 6 + 0}&{9 + 18 + 11} \end{array}} \right)$
$\Rightarrow \left( {\begin{array}{*{20}{c}} {27}&{ - 1}&1 \\ {45}&{18}&{ - 5} \\ 5&{10}&{28} \end{array}} \right)$
Hence we get,
D) ${{\text{A}}^{\text{2}}}{\text{ + 6A + 11I}} = \left( {\begin{array}{*{20}{c}} {27}&{ - 1}&1 \\ {45}&{18}&{ - 5} \\ 5&{10}&{28} \end{array}} \right)$
Hence matrix value of ${{\text{A}}^{{\text{ - 1}}}}$ is equal to the expression of option A,
That is, ${{\text{A}}^{{\text{ - 1}}}} = {{\text{A}}^{\text{2}}}{\text{ - 6A + 11I = }}\left( {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\ { - 15}&6&{ - 5} \\ 5&{ - 2}&2 \end{array}} \right)$

$\therefore$ Option A is the correct answer.

Note: We have to focus while using an identity matrix to put values correctly and also we have to concentrate on Multiply the First matrix and second matrix of the given matrices for ${{\text{A}}^{\text{2}}}$ . Next to find the answer, while multiple and adding the matrices enter the value correctly.