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Using Elementary transformation, find the inverse each of matrices, if it exists in ques 1 to 17.
$\left[\begin{array}{ll}2 & -6 \\ 1 & -2\end{array}\right]$
$\left[\begin{array}{ll}2 & -6 \\ 1 & -2\end{array}\right]$
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Verified Answer
Let $A=\left[\begin{array}{ll}2 & -6 \\ 1 & -2\end{array}\right]$
We know that $\mathrm{A}=\mathrm{IA}$
$$
\begin{aligned}
&\Rightarrow\left[\begin{array}{ll}
2 & -6 \\
1 & -2
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right] \mathrm{A} \\
&\Rightarrow\left[\begin{array}{cc}
1 & -4 \\
1 & -2
\end{array}\right]=\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right] \mathrm{A}, \mathrm{R}_1 \Rightarrow \mathrm{R}_1-\mathrm{R}_2 \\
&\Rightarrow\left[\begin{array}{cc}
1 & -4 \\
0 & 2
\end{array}\right]=\left[\begin{array}{cc}
1 & -1 \\
-1 & 2
\end{array}\right] \mathrm{A}, \mathrm{R}_2 \Rightarrow \mathrm{R}_2-\mathrm{R}_1 \\
&\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right]=\left[\begin{array}{cc}
-1 & 3 \\
-1 & 2
\end{array}\right] \mathrm{A}, \mathrm{R}_1 \Rightarrow \mathrm{R}_1+2 \mathrm{R}_2 \\
&\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
-1 & 3 \\
\frac{-1}{2} & 1
\end{array}\right] \mathrm{A}, \mathrm{R}_1 \Rightarrow \frac{1}{2} \mathrm{R}_1
\end{aligned}
$$
Hence, $A^{-1}=\left[\begin{array}{cc}-1 & 3 \\ \frac{-1}{2} & 1\end{array}\right]$
We know that $\mathrm{A}=\mathrm{IA}$
$$
\begin{aligned}
&\Rightarrow\left[\begin{array}{ll}
2 & -6 \\
1 & -2
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right] \mathrm{A} \\
&\Rightarrow\left[\begin{array}{cc}
1 & -4 \\
1 & -2
\end{array}\right]=\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right] \mathrm{A}, \mathrm{R}_1 \Rightarrow \mathrm{R}_1-\mathrm{R}_2 \\
&\Rightarrow\left[\begin{array}{cc}
1 & -4 \\
0 & 2
\end{array}\right]=\left[\begin{array}{cc}
1 & -1 \\
-1 & 2
\end{array}\right] \mathrm{A}, \mathrm{R}_2 \Rightarrow \mathrm{R}_2-\mathrm{R}_1 \\
&\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right]=\left[\begin{array}{cc}
-1 & 3 \\
-1 & 2
\end{array}\right] \mathrm{A}, \mathrm{R}_1 \Rightarrow \mathrm{R}_1+2 \mathrm{R}_2 \\
&\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
-1 & 3 \\
\frac{-1}{2} & 1
\end{array}\right] \mathrm{A}, \mathrm{R}_1 \Rightarrow \frac{1}{2} \mathrm{R}_1
\end{aligned}
$$
Hence, $A^{-1}=\left[\begin{array}{cc}-1 & 3 \\ \frac{-1}{2} & 1\end{array}\right]$
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