If $A^{-1}=\frac{-1}{2}\left[\begin{array}{cc}5 & 8 \\ -1 & 2\end{array}\right]$, then $2 A+I_2=$, where…
If $A^{-1}=\frac{-1}{2}\left[\begin{array}{cc}5 & 8 \\ -1 & 2\end{array}\right]$, then $2 A+I_2=$, where $I_2$ is a unit matrix of order 2
- $\left[\begin{array}{ll}5 & 8 \\ 1 & 2\end{array}\right]$
- $\left[\begin{array}{ll}5 & 8 \\ 2 & 2\end{array}\right]$
- $\left[\begin{array}{ll}2 & 4 \\ 1 & 1\end{array}\right]$
- $\left[\begin{array}{ll}5 & 8 \\ 2 & 3\end{array}\right]$
Solution
$\begin{aligned}
& \mathrm{A}^{-1}=\frac{-1}{2}\left[\begin{array}{cc}
1 & -4 \\
-1 & 2
\end{array}\right]=\left[\begin{array}{cc}
\frac{-1}{2} & 2 \\
\frac{1}{2} & -1
\end{array}\right] \\
& \mathrm{AA}^{-1}=\mathrm{I} \Rightarrow \mathrm{A}\left[\begin{array}{cc}
\frac{-1}{2} & 2 \\
\frac{1}{2} & -1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
\end{aligned}$
$\begin{aligned}
& \mathrm{R}_2 \rightarrow \mathrm{R}_1+\mathrm{R}_2 \\
& \mathrm{~A}\left[\begin{array}{ll}
\frac{-1}{2} & 2 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \\
& \mathrm{R}_1 \rightarrow \mathrm{R}_1+2 \mathrm{R}_2 \\
& \mathrm{~A}\left[\begin{array}{cc}
\frac{-1}{2} & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right] \\
& \mathrm{R}_1 \rightarrow-2 \mathrm{R}_1 \\
& \mathrm{~A}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
2 & 4 \\
1 & 1
\end{array}\right] \Rightarrow \mathrm{A}=\left[\begin{array}{ll}
2 & 4 \\
1 & 1
\end{array}\right] \\
& \therefore 2 \mathrm{~A}+\mathrm{I}_2=\left[\begin{array}{ll}
4 & 8 \\
2 & 2
\end{array}\right]+\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
5 & 8 \\
2 & 3
\end{array}\right]
\end{aligned}$
Asked in: MHT CET 2021 (21 Sep Shift 1)
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