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REPEATER 2024 — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsREPEATER 20242 Marks
Question
Find the inverse of matrix $A$ by elementary row transformation, where $A=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$.
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Answer

$A.A^{-1}=I$
$\therefore \quad\left[\begin{array}{cc}2 & -3 \\-1 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
Using $R_1 \rightarrow R_1+R_2$
$\begin{array}{l}\therefore \quad\left[\begin{array}{cc} \ \ \ 1\   \ \ - \ 1 \\-1 \ \ \ \ \ \ 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\end{array}$
 Using $ R_2 \rightarrow R_2+R_1$
$\therefore \quad\left[\begin{array}{cc}1 & -1 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 1 \\1 & 2
\end{array}\right]$
Using $R_1 \rightarrow R_1+R_2$
$\begin{array}{lrl}\therefore & {\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]}\end{array} $
$ \therefore  A^{-1}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

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