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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]2 Marks
Question
If $A=\left[\begin{array}{ll}2 & 4 \\ 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ then find $\left(A^{-1} B^{-1}\right)$
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Answer

$\begin{array}{l}\left(A^{-1} B^{-1}\right)=(B A)^{-1} \ldots . .\left[\because(A B)^{-1}=B^{-1} A^{-1}\right] \end{array}$
$ \therefore A=\left[\begin{array}{ll}2 & 4 \\ 1 & 3\end{array}\right] $
$ A=(2 \times 3)-(4 \times 1) $
$ A=6-4 \\ A=2 $
$ \therefore B=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] $
$ B=(1 \times 1)-(1 \times 0)$
$ B=1-0$ 
$B=1$
$\left(A^{-1} B^{-1}\right)=\frac{1}{|A|}(\operatorname{adj} A) \cdot \frac{1}{|B|}(\operatorname{adj} B)$
$ =\frac{1}{2}\left[\begin{array}{cc}3 & -4 \\ -1 & 2\end{array}\right] \times \frac{1}{1}\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] $
$ =\frac{1}{2}\left[\begin{array}{cc}3 & -4 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \\ =\frac{1}{2}\left[\begin{array}{cc}3 & -7 \\ -1 & 3\end{array}\right]$

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