Determinants — MATHS STD 12 Science — Question

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&1&3 \\ 4&{ - 1}&0 \\ { - 7}&2&1 \end{array}} \right| = 2\left\{ {\left( { - 1} \right) - \left( 4 \right)} \right\} + 3\left( {8 - 7} \right) = - 3 \ne 0$

$\therefore {A^{ - 1}}$ exists.

${A_{11}} = + \left| {\begin{array}{*{20}{c}} { - 1}&0 \\ 2&1 \end{array}} \right| = + \left( { - 1 - 0} \right) = - 1,{A_{12}} = - \left| {\begin{array}{*{20}{c}} 4&0 \\ { - 7}&1 \end{array}} \right| = - \left( {4 - 0} \right) = - 4$

${A_{13}} = + \left| {\begin{array}{*{20}{c}} 4&{ - 1} \\ { - 7}&2 \end{array}} \right| = + \left( {8 - 7} \right) = 1,{A_{21}} = - \left| {\begin{array}{*{20}{c}} 1&3 \\ 2&1 \end{array}} \right| = - \left( {1 - 6} \right) = 5$

${A_{22}} = + \left| {\begin{array}{*{20}{c}} 2&3 \\ { - 7}&1 \end{array}} \right| = + \left( {2 + 21} \right) = 23,{A_{23}} = - \left| {\begin{array}{*{20}{c}} 2&1 \\ { - 7}&2 \end{array}} \right| = - \left( {4 + 7} \right) = 11$

${A_{31}} = + \left| {\begin{array}{*{20}{c}} 1&3 \\ { - 1}&0 \end{array}} \right| = + \left( {0 + 3} \right) = 3,{A_{32}} = - \left| {\begin{array}{*{20}{c}} 2&3 \\ 4&0 \end{array}} \right| = - \left( {0 - 12} \right) = 12$

${A_{33}} = + \left| {\begin{array}{*{20}{c}} 2&1 \\ 4&{ - 1} \end{array}} \right| = + \left( { - 2 - 4} \right) = - 6$

$\therefore adj.A = \left[ {\begin{array}{*{20}{c}} { - 1}&4&1 \\ 5&{23}&{ - 11} \\ 3&{12}&{ - 6} \end{array}} \right]' = \left[ {\begin{array}{*{20}{c}} { - 1}&4&1 \\ 5&{23}&{ - 11} \\ 3&{12}&{ - 6} \end{array}} \right]$

$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A = \frac{{ - 1}}{3} \left[ {\begin{array}{*{20}{c}} { - 1}&5&2 \\ { - 4}&{23}&{12} \\ 1&{ - 11}&{ - 6} \end{array}} \right]$