Determinants — Maths STD 12 Science — Question

|AB| = 14 -25 = -11 

$\Rightarrow$ (AB) is non singular and hence, (AB)^-1 exists

$\therefore$ (AB)^-1= $\frac{1}{{|AB|}}adj(AB) = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} { - 14}&{ - 5} \\ { - 5}&{ - 1} \end{array}} \right]$
$$Now, |A| = - 8 - 3 = - 11

 $\Rightarrow$ A is non singular and hence A^-1 exists

Also, |B| = 3 - 2 = 1 

$\Rightarrow$ B is non- singular and hence B^-1 exists

A^-1 = $\frac{{ - 1}}{{11}}\left[ {\begin{array}{*{20}{c}} { - 4}&{ - 3} \\ { - 1}&2 \end{array}} \right]$

B^-1= $\frac{1}{1}\left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]$

B^-1A^-1 = $\frac{{ - 1}}{{11}}\left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 4}&{ - 3} \\ { - 1}&2 \end{array}} \right]$

= $\frac{{ - 1}}{{11}}\left[ {\begin{array}{*{20}{c}} { - 14}&{ - 5} \\ { - 5}&{ - 1} \end{array}} \right]$

Hence, (AB)^-1 = B^-1A^-1