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Using Elementary transformation, find the inverse each of matrices, if it exists in ques 1 to 17.
$\left[\begin{array}{cc}6 & -3 \\ -2 & 1\end{array}\right]$
$\left[\begin{array}{cc}6 & -3 \\ -2 & 1\end{array}\right]$
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Verified Answer
et $A=\left[\begin{array}{cc}6 & -3 \\ -2 & 1\end{array}\right]$
We know that $\mathrm{A}=\mathrm{IA}$
$$
\begin{aligned}
\Rightarrow & {\left[\begin{array}{cc}
6 & -3 \\
-2 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right] \mathrm{A} } \\
\Rightarrow & {\left[\begin{array}{cc}
0 & 0 \\
-2 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & -3 \\
0 & 1
\end{array}\right] \mathrm{A}, \mathrm{R}_1 \Rightarrow \mathrm{R}_1-3 \mathrm{R}_2 }
\end{aligned}
$$
As we get all zeroes in the first row of L.H.S.
$\therefore$ The inverse of the matrix does not exist.
We know that $\mathrm{A}=\mathrm{IA}$
$$
\begin{aligned}
\Rightarrow & {\left[\begin{array}{cc}
6 & -3 \\
-2 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right] \mathrm{A} } \\
\Rightarrow & {\left[\begin{array}{cc}
0 & 0 \\
-2 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & -3 \\
0 & 1
\end{array}\right] \mathrm{A}, \mathrm{R}_1 \Rightarrow \mathrm{R}_1-3 \mathrm{R}_2 }
\end{aligned}
$$
As we get all zeroes in the first row of L.H.S.
$\therefore$ The inverse of the matrix does not exist.
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