Determinants — Maths STD 12 Science — Question

$\therefore |A| = \left| {\begin{array}{*{20}{c}} 1&2&3 \\ 0&2&4 \\ 0&0&5 \end{array}} \right| $ = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) $ = 10 \ne 0$

$\therefore {A^{ - 1}}$ exists.

${A_{11}} = + \left[ {\begin{array}{*{20}{c}} 2&4 \\ 0&5 \end{array}} \right] = + \left( {10 - 0} \right) = 10,$ ${A_{12}} = - \left| {\begin{array}{*{20}{c}} 0&4 \\ 0&5 \end{array}} \right| = - \left( {0 - 0} \right) = 0$

${A_{13}} = + \left[ {\begin{array}{*{20}{c}} 0&2 \\ 0&0 \end{array}} \right] = + \left( {0 - 0} \right) = 0,$ ${A_{21}} = - \left| {\begin{array}{*{20}{c}} 2&3 \\ 0&5 \end{array}} \right| = - \left( {10 - 0} \right) = - 10$

${A_{22}} = + \left| {\begin{array}{*{20}{c}} 1&3 \\ 0&5 \end{array}} \right| = + \left( {5 - 0} \right) = 5,$ ${A_{23}} = - \left| {\begin{array}{*{20}{c}} 1&2 \\ 0&0 \end{array}} \right| = - \left( {0 - 0} \right) = 0$

${A_{31}} = + \left| {\begin{array}{*{20}{c}} 2&3 \\ 2&4 \end{array}} \right| = + \left( {8 - 6} \right) = 2,$ ${A_{32}} = - \left| {\begin{array}{*{20}{c}} 1&3 \\ 0&4 \end{array}} \right| = - \left( {4 - 0} \right) = - 4$

${A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&2 \\ 0&2 \end{array}} \right| = + \left( {2 - 0} \right) = 2$

$\therefore adj.A = \left| {\begin{array}{*{20}{c}} {10}&0&0 \\ { - 10}&5&0 \\ 2&{ - 4}&2 \end{array}} \right|'$

$= \left| {\begin{array}{*{20}{c}} {10}&{ - 10}&2 \\ 0&5&{ - 4} \\ 0&0&2 \end{array}} \right|$

$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A=\frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {10}&{ - 10}&2 \\ 0&5&{ - 4} \\ 0&0&2 \end{array}} \right]$