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# Find ${A^{ - 1}}$ if $A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ { - 2}&1&3 \\ 4&{ - 2}&2 \end{array}} \right]$

Last updated date: 13th Jun 2024
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Hint: We have given a $3 \times 3$ matrix we have to calculate its inverse. So firstly we have to calculate the determinant of the matrix if the determinant is equal to zero. Then the inverse of the matrix does not exist. If the determinant is not equal to zero then the inverse of the matrix exists. After we calculate the cofactor of each row and then we calculate the adjoint of $A$. Adjoint of $A$ can be calculated by putting the co-factors in a matrix and taking its transpose then we apply formula of inverse of the matrix
Inverse of $A = {A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}$
Here $adjA$ is adjoint of $A$ and $\left| A \right|$ is determinant of $A$.

We have been given a matrix
$A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ { - 2}&1&3 \\ 4&{ - 2}&2 \end{array}} \right]$
we calculate its determinant,
$\left| A \right| = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ { - 2}&1&3 \\ 4&{ - 2}&2 \end{array}} \right]$
$= 1(2 + 6) + 1( - 4 - 12) + 2(4 - 4)$
$= 1(8) + 1( - 16) + 2(0)$
$\Rightarrow 8 - 16 + 0 = - 8$
$\left| A \right| = - 8 \ne 0$
So, ${A^{ - 1}}$ exist
Now we calculate cofactors of Matrix $A$
Cofactor of first row is given as :
$\left| {\begin{array}{*{20}{c}} 1&3 \\ { - 2}&2 \end{array}} \right|, - \left| {\begin{array}{*{20}{c}} { - 2}&3 \\ 4&2 \end{array}} \right|,\left| {\begin{array}{*{20}{c}} { - 2}&1 \\ 4&{ - 2} \end{array}} \right| = - 8,16,0$
Cofactor of second row is given as:
$- \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ { - 2}&2 \end{array}} \right|,\left| {\begin{array}{*{20}{c}} 1&2 \\ 4&2 \end{array}} \right|, - \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 4&{ - 2} \end{array}} \right| = - 2, - 6, - 2$
Cofactor of Third row is given as:
$\left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 1&3 \end{array}} \right|, - \left| {\begin{array}{*{20}{c}} 1&2 \\ { - 2}&3 \end{array}} \right|,\left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ { - 2}&1 \end{array}} \right| = - 5, - 7, - 1$
Now, $adjA = {\left[ {\begin{array}{*{20}{c}} 8&{16}&0 \\ { - 2}&{ - 6}&{ - 2} \\ { - 5}&{ - 7}&{ - 1} \end{array}} \right]^1} = \left[ {\begin{array}{*{20}{c}} 8&{ - 2}&{ - 5} \\ {16}&{ - 6}&{ - 7} \\ 0&{ - 2}&{ - 1} \end{array}} \right]$
Adjoint of $A = adjA = \left[ {\begin{array}{*{20}{c}} 8&{ - 2}&{ - 5} \\ {16}&{ - 6}&{ - 7} \\ 0&{ - 2}&{ - 1} \end{array}} \right]$

${A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}} = \dfrac{1}{{ - 8}}\left[ {\begin{array}{*{20}{c}} 8&{ - 2}&{ - 5} \\ {16}&{ - 6}&{ - 7} \\ 0&{ - 2}&{ - 1} \end{array}} \right]$

${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} { - 1}&\dfrac{1}{{4}}&\dfrac{5}{{ 8}} \\ { - 2}&\dfrac{3}{{4}}&\dfrac{7}{{8}} \\ 0&\dfrac{1}{{4}}&\dfrac{1}{{ 8}} \end{array}} \right]$

Note: In mathematics a matrix is a rectangular array or table of numbers symbols, or expressions, arranged in row and column. Matrices with $m$ rows and $n$ columns are called $m \times n$ matrices. Matrices can be used to compactly write and work with multiple linear questions, that is a system of linear equations. A square matrix has an inverse if its determinant is not equal to zero. Such matrices are also called invertible matrices.