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If $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$, then $A^{-1}=$
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1436 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{5}(A-4 h)$
We have,
$$
\begin{aligned}
A & =\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right] \\
|A| & =1(1-4)-2(2-4)+2(4-2=-3+4+4=5 \\
\operatorname{Adj} A & =\left[\begin{array}{ccc}
-3 & 2 & 2 \\
2 & -3 & 2 \\
2 & 2 & -3
\end{array}\right] \\
A^{-1} & =\frac{1}{|A|} \text { Adj A }=\frac{1}{5}\left[\begin{array}{ccc}
-3 & 2 & 2 \\
2 & -3 & 2 \\
2 & 2 & -3
\end{array}\right] \\
A^{-1} & =\frac{1}{5}\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-\frac{1}{5}\left[\begin{array}{ccc}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{array}\right] \\
A^{-1} & =\frac{1}{5}\left[\begin{array}{ll}
A-4 I
\end{array}\right]
\end{aligned}
$$
$$
\begin{aligned}
A & =\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right] \\
|A| & =1(1-4)-2(2-4)+2(4-2=-3+4+4=5 \\
\operatorname{Adj} A & =\left[\begin{array}{ccc}
-3 & 2 & 2 \\
2 & -3 & 2 \\
2 & 2 & -3
\end{array}\right] \\
A^{-1} & =\frac{1}{|A|} \text { Adj A }=\frac{1}{5}\left[\begin{array}{ccc}
-3 & 2 & 2 \\
2 & -3 & 2 \\
2 & 2 & -3
\end{array}\right] \\
A^{-1} & =\frac{1}{5}\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-\frac{1}{5}\left[\begin{array}{ccc}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{array}\right] \\
A^{-1} & =\frac{1}{5}\left[\begin{array}{ll}
A-4 I
\end{array}\right]
\end{aligned}
$$
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