Determinants — Maths STD 12 Science — Question

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 3&7 \\ 2&5 \end{array}} \right| = 15 - 14 = 1 \ne 0$

$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A $ $= \frac{1}{1}\left[ {\begin{array}{*{20}{c}} 5&{ - 7} \\ { - 2}&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5&{ - 7} \\ { - 2}&3 \end{array}} \right]$

Matrix $B = \left[ {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right]$

$\therefore \left| B \right| = \left| {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right| = 54 - 56 = - 2 \ne 0$

$\therefore {B^{ - 1}} = \frac{1}{{\left| B \right|}}adj.B = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} 9&{ - 8} \\ { - 7}&6 \end{array}} \right]$

Now $AB = \left[ {\begin{array}{*{20}{c}} 3&7 \\ 2&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} {18 + 49}&{24 + 63} \\ {12 + 35}&{16 + 45} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {67}&{87} \\ {47}&{61} \end{array}} \right]$

$\therefore \left| {AB} \right| = \left| {\begin{array}{*{20}{c}} {67}&{87} \\ {47}&{61} \end{array}} \right| $ = 67(61) - 87(47) = 4087 - 4089 $= - 2 \ne 0$

Now L.H.S. = ${\left( {AB} \right)^{ - 1}} = \frac{1}{{\left| {AB} \right|}}adj.\left( {AB} \right) $ $= \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} {61}&{ - 87} \\ { - 47}&{67} \end{array}} \right]$….(i)

R.H.S. = B^-1A^-1 $ = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} 9&{ - 8} \\ { - 7}&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&{ - 7} \\ { - 2}&3 \end{array}} \right]$

$= \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} {45 + 16}&{ - 63 - 24} \\ { - 35 - 12}&{49 + 18} \end{array}} \right]$

$= \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} {61}&{ - 87} \\ { - 47}&{67} \end{array}} \right]$….(ii)

$\therefore$ From eq. (i) and (ii), we get

L.H.S. = R.H.S.

$ \Rightarrow {\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$