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If the matrix $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ satisfies the matrix equation $\mathrm{A}^2-4 \mathrm{~A}-5 \mathrm{I}=0$, then $\mathrm{A}^{-1}=$
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Verified Answer
The correct answer is:
$\frac{1}{5}\left[\begin{array}{ccc}-3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3\end{array}\right]$
Given matrix $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$.
Multiply both sides by $\mathrm{A}^{-1}$.
$\begin{aligned}
& \left(\mathrm{A}^2 \mathrm{~A}^{-1}\right)-4\left(\mathrm{~A} \cdot \mathrm{A}^{-1}\right)-5\left(\mathrm{I} \cdot \mathrm{A}^{-1}\right)=0 . \mathrm{A}^{-1} \\
& \mathrm{~A}-4 \mathrm{I}-5 \mathrm{~A}^{-1}=0 \\
& 5 \mathrm{~A}^{-1}=\mathrm{A}-4 \mathrm{I} \\
& \mathrm{A}^{-1}=\frac{1}{5}(\mathrm{~A}-4 \mathrm{I}) .
\end{aligned}
$Put the value of A,
$\begin{aligned}
& \mathrm{A}^{-1}=\frac{1}{5}\left(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-4\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right) \\
& \mathrm{A}^{-1}=\frac{1}{5}\left(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-\left[\begin{array}{lll}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{array}\right]\right) \\
& \text { So, } \mathrm{A}^{-1}=\frac{1}{5}\left[\begin{array}{ccc}
-3 & 2 & 2 \\
2 & -3 & 2 \\
2 & 2 & -3
\end{array}\right]
\end{aligned}$
Multiply both sides by $\mathrm{A}^{-1}$.
$\begin{aligned}
& \left(\mathrm{A}^2 \mathrm{~A}^{-1}\right)-4\left(\mathrm{~A} \cdot \mathrm{A}^{-1}\right)-5\left(\mathrm{I} \cdot \mathrm{A}^{-1}\right)=0 . \mathrm{A}^{-1} \\
& \mathrm{~A}-4 \mathrm{I}-5 \mathrm{~A}^{-1}=0 \\
& 5 \mathrm{~A}^{-1}=\mathrm{A}-4 \mathrm{I} \\
& \mathrm{A}^{-1}=\frac{1}{5}(\mathrm{~A}-4 \mathrm{I}) .
\end{aligned}
$Put the value of A,
$\begin{aligned}
& \mathrm{A}^{-1}=\frac{1}{5}\left(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-4\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right) \\
& \mathrm{A}^{-1}=\frac{1}{5}\left(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-\left[\begin{array}{lll}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{array}\right]\right) \\
& \text { So, } \mathrm{A}^{-1}=\frac{1}{5}\left[\begin{array}{ccc}
-3 & 2 & 2 \\
2 & -3 & 2 \\
2 & 2 & -3
\end{array}\right]
\end{aligned}$
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