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If $A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$, then $A \cdot \operatorname{adj}(A)$ is equal to
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Verified Answer
The correct answer is:
$\left[\begin{array}{lll}8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8\end{array}\right]$
We have,
$$
\begin{aligned}
A &=\left[\begin{array}{ccc}
1 & -2 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right] \\
\therefore \quad|A| &=\left[\begin{array}{ccc}
1 & -2 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right] \\
&=1[8-6]+3[6-4]=2+6=8
\end{aligned}
$$
Now, we know that
$$
A \cdot(\operatorname{adj} A)=8 l=\left[\begin{array}{lll}
8 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 8
\end{array}\right]
$$
$$
\begin{aligned}
A &=\left[\begin{array}{ccc}
1 & -2 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right] \\
\therefore \quad|A| &=\left[\begin{array}{ccc}
1 & -2 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right] \\
&=1[8-6]+3[6-4]=2+6=8
\end{aligned}
$$
Now, we know that
$$
A \cdot(\operatorname{adj} A)=8 l=\left[\begin{array}{lll}
8 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 8
\end{array}\right]
$$
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