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DETERMINANTS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDETERMINANTS5 Marks
Question
$\text{Let A} = \begin{bmatrix}3&7\\2&5\end{bmatrix}\text{and B} = \begin{bmatrix}6&8\\7&9\end{bmatrix}.$Verify that $(AB)^{-1} = B^{-1}A^{-1}.$
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Answer

$\text{Given}:\ \text{Matrix A}=\begin{bmatrix}3&7\\2&5\end{bmatrix}$ 
$\therefore\ \text{|A|}=\begin{vmatrix}3&7\\2&5\end{vmatrix}=15-14=1\neq0$
$\therefore\ \text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj. A}=\frac{1}{1}\begin{bmatrix}5&-7\\-2&3\end{bmatrix}=\begin{bmatrix}5&-7\\-2&3\end{bmatrix}$
$\text{Matrix B}=\begin{bmatrix}6&8\\7&9\end{bmatrix}$ $\therefore\ \text{|B|}=\begin{vmatrix}6&8\\7&9\end{vmatrix}=54-56=-2\neq0$
$\therefore\ \text{B}^{-1}=\frac{1}{\text{|B|}}\text{adj. B}=\frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}$
$\text{Now}\ \text{AB}=\begin{bmatrix}3&7\\2&5\end{bmatrix}\begin{bmatrix}6&8\\7&9\end{bmatrix}=\begin{bmatrix}18+49&24+63\\12+35&16+45\end{bmatrix}=\begin{bmatrix}67&87\\47&61\end{bmatrix}$
$\therefore\ \text{|AB|}=\begin{vmatrix}67&87\\47&61\end{vmatrix}=67(61)-87(47)=4087-4089=-2\neq0$
$\text{Now}\ \text{L.H.S.}=\left(\text{AB}\right)^{-1}=\frac{1}{\text{|AB|}}\text{adj. (AB)}=\frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} \dots\dots(1)$
$\text{R.H.S.}=\text{B}^{-1}\text{A}^{-1}=\frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}\begin{bmatrix}5&-7\\-2&3\end{bmatrix}=\frac{1}{-2}\begin{bmatrix}45+16&-63-24\\-35-12&49+18\end{bmatrix}$
$=\frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} \dots\dots(2)$
From eq. $(1)$ and $(2),$ we get $\text{L.H.S. = R.H.S.}$
$\Rightarrow (AB)^{-1} = B^{-1}A^{-1}$

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