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Find the inverse of the matrix (if it exists) given $$\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right]$$
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## QuestionMathsClass 12

Find the inverse of the matrix (if it exists) given $$\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right]$$

## Answer

Let $$A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right]$$
$$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right| = 1\left( {8 - 6} \right) - \left( { - 1} \right)\left( {0 + 9} \right) + 2\left( {0 - 6} \right) = - 1 \ne 0$$
$$\therefore {A^{ - 1}}$$ exists.
$${A_{11}} = + \left| {\begin{array}{*{20}{c}} 2&{ - 3} \\ { - 2}&4 \end{array}} \right| = + \left( {8 - 6} \right) = 2,{A_{12}} = - \left| {\begin{array}{*{20}{c}} 0&{ - 3} \\ 3&4 \end{array}} \right| = - \left( {0 + 9} \right) = - 9$$
$${A_{13}} = + \left| {\begin{array}{*{20}{c}} 0&2 \\ 3&{ - 2} \end{array}} \right| = + \left( {0 - 6} \right) = 6,{A_{21}} = - \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ { - 2}&4 \end{array}} \right| = - \left( { - 4 + 4} \right) = 0$$
$${A_{22}} = + \left| {\begin{array}{*{20}{c}} 2&2 \\ 3&4 \end{array}} \right| = + \left( {4 - 6} \right) = - 2,{A_{23}} = - \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 3&{ - 2} \end{array}} \right| = - \left( { - 2 + 3} \right) = - 1$$
$${A_{31}} = + \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 2&3 \end{array}} \right| = + \left( {3 - 4} \right) = - 1,{A_{32}} = - \left| {\begin{array}{*{20}{c}} 1&2 \\ 0&3 \end{array}} \right| = - \left( { - 3 - 0} \right) = 3$$
$${A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 0&2 \end{array}} \right| = + \left( {2 - 0} \right) = 2,$$
$$\therefore adj.A = \left[ {\begin{array}{*{20}{c}} 2&{ - 9}&{ - 6} \\ 0&{ - 2}&{ - 1} \\ { - 1}&3&2 \end{array}} \right]' = \left[ {\begin{array}{*{20}{c}} 2&0&{ - 1} \\ { - 9}&{ - 2}&3 \\ { - 6}&{ - 1}&2 \end{array}} \right]$$
$$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A = \frac{{ - 1}}{1} \left[ {\begin{array}{*{20}{c}} 2&0&{ - 1} \\ { - 9}&{ - 2}&3 \\ { - 6}&{ - 1}&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2}&0&1 \\ 9&2&{ - 3} \\ 6&1&{ - 2} \end{array}} \right]$$
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