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Using Elementary transformation, find the inverse each of matrices, if it exists in ques 1 to 17.
$\left[\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right]$
$\left[\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right]$
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Verified Answer
We know that $\mathrm{A}=\mathrm{IA}$
$$
\begin{aligned}
&{\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \mathrm{A}} \\
&\Rightarrow\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-2 & 1
\end{array}\right] \mathrm{A}, \mathrm{R}_2 \Rightarrow \mathrm{R}_2-2 \mathrm{R}_1 \\
&\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & -3 \\
-2 & 1
\end{array}\right] \mathrm{A}, \mathrm{R}_1 \Rightarrow \mathrm{R}-3 \mathrm{R}_2
\end{aligned}
$$
Hence, $\mathrm{A}^{-1}=\left[\begin{array}{cc}7 & -3 \\ -2 & 1\end{array}\right]$
$$
\begin{aligned}
&{\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \mathrm{A}} \\
&\Rightarrow\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-2 & 1
\end{array}\right] \mathrm{A}, \mathrm{R}_2 \Rightarrow \mathrm{R}_2-2 \mathrm{R}_1 \\
&\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & -3 \\
-2 & 1
\end{array}\right] \mathrm{A}, \mathrm{R}_1 \Rightarrow \mathrm{R}-3 \mathrm{R}_2
\end{aligned}
$$
Hence, $\mathrm{A}^{-1}=\left[\begin{array}{cc}7 & -3 \\ -2 & 1\end{array}\right]$
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