Determinants — Maths STD 12 Science — Question

$\therefore \left| A \right| = \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ 4&3 \end{array}} \right] = 6 - \left( { - 8} \right) = 6 + 8 = 14 \ne 0$

$\therefore$ Matrix A is non-singular and hence ${A^{ - 1}}$ exist.

Now adj. A $ = \left[ {\begin{array}{*{20}{c}} 3&2 \\ { - 4}&2 \end{array}} \right]$ And ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A = \frac{1}{{14}}\left[ {\begin{array}{*{20}{c}} 3&2 \\ { - 4}&2 \end{array}} \right]$