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

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]2 Marks
Question
If $A=\left[\begin{array}{ll}2 & 0 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 2\end{array}\right]$ then find the matrix $X$ such that $A^{-1} X=B$
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Answer

$|A|=2 \\ A_{11}=(-1)^{1+1} M_{11}=M_{11}=1 $
$A_{12}=(-1)^{1+2} M_{12}=-M_{12}=0 $
$ A_{21}=(-1)^{2+1} M_{21}=-M_{21}=0 $
$ A_{22}=(-1)^{2+2} M_{22}=M_{22}=2$
$ \operatorname{adj}|A|=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]^{ T } $
$ =\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{|A|}\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right] $
$ =\frac{1}{2}\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]$
Let $X =\left[\begin{array}{l} a \\ b \end{array}\right] $
$ \therefore A ^{-1}= B $
$ \Rightarrow \frac{1}{2}\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]\left[\begin{array}{l} a \\ b \end{array}\right]=\left[\begin{array}{l}1 \\ 2\end{array}\right] $
$ \Rightarrow \frac{1}{2}\left[\begin{array}{l} a \\ 2 b \end{array}\right]=\left[\begin{array}{l}1 \\ 2\end{array}\right] $
$ \Rightarrow \frac{ a }{2}=1 \text { and } b =2 $
$ \Rightarrow a =2 \text { an } b =2 $
$ \therefore X =\left[\begin{array}{l}2 \\ 2\end{array}\right]$

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