If $A=\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]$ and $A^{2}-5 A-6 I=0$, then $A^{-1}=$
If $A=\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]$ and $A^{2}-5 A-6 I=0$, then $A^{-1}=$
- $\frac{1}{6}\left[\begin{array}{cc}-1 & 5 \\ 2 & 4\end{array}\right]$
- $\frac{1}{6}\left[\begin{array}{cc}-1 & 5 \\ -2 & -4\end{array}\right]$
- $\frac{1}{6}\left[\begin{array}{cc}-1 & 5 \\ 2 & -4\end{array}\right]$
- $\frac{1}{6}\left[\begin{array}{cc}1 & 5 \\ 2 & -4\end{array}\right]$
Solution
$\begin{array}{l}
A=\left[\begin{array}{cc}
4 & 5 \\
2 & 1
\end{array}\right] \Rightarrow|A|=4-10--6 \text { and }(\operatorname{adj} A)-\left[\begin{array}{cc}
1 & 5 \\
-2 & 4
\end{array}\right] \\
\therefore A^{-1}=-\frac{1}{6}\left[\begin{array}{cc}
1 & -5 \\
-2 & 4
\end{array}\right]=\frac{1}{6}\left[\begin{array}{cc}
-1 & 5 \\
2 & -4
\end{array}\right]
\end{array}$
Asked in: MHT CET 2020 (19 Oct Shift 2)
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