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

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]2 Marks
Question
Find $A ^{-1}$ using column transformations: $A=\left[\begin{array}{cc}5 & 3 \\3 & -2\end{array}\right]$
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Answer

We know that $A A^{-1}=1$
$\therefore\left[\begin{array}{cc}5 & 3 \\3 & -2\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Applying $C_1 \rightarrow 2 C_2-C_1$, we get
$\left[\begin{array}{cc}1 & 3 \\-7 & -2\end{array}\right] A^{-1}=\left[\begin{array}{cc}-1 & 0 \\ 2 & 1 \end{array}\right]$
Applying $C_2 \rightarrow C_2-3 C_1$, we get
$\left[\begin{array}{cc}1 & 0 \\-7 & 19\end{array}\right] A^{-1}=\left[\begin{array}{cc}-1 & 3 \\2 & -5\end{array}\right]$
Applying $C_2 \rightarrow\left(\frac{1}{19}\right) C_2$, we get
$\left[\begin{array}{cc}1 & 0 \\-7 & 1\end{array}\right] A^{-1}=\left[\begin{array}{cc}-1 & \frac{3}{19} \\2 & \frac{-5}{19}\end{array}\right]$
Applying $C_1 \rightarrow C_1+7 C_2$, we get
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] A ^{-1}=\left[\begin{array}{cc}\frac{2}{19} & \frac{3}{19} \\\frac{3}{19}& \frac{-5}{19}\end{array}\right]} \end{array}$
$\therefore A ^{-1}=\frac{1}{19}\left[\begin{array}{cc}2 & 3 \\3 & -5\end{array}\right]$
 

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