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If $A=\left[\begin{array}{ll}3 & 4 \\ 5 & 7\end{array}\right]$, then $\mathrm{A}$. ( $\operatorname{adj} A$ ) is equal to
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Verified Answer
The correct answer is:
$|A|$
$\mathrm{A}=\left[\begin{array}{ll}3 & 4 \\ 5 & 7\end{array}\right]$
$|\mathrm{A}|=21-20=1$
$\begin{array}{l}
\therefore \quad \mathrm{A}(\operatorname{adj} \mathrm{A})=\left[\begin{array}{ll}
3 & 4 \\
5 & 7
\end{array}\right]\left[\begin{array}{cc}
7 & -4 \\
-5 & 3
\end{array}\right] \\
=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=1 .\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=|\mathrm{A}| . \mathrm{I}
\end{array}$
$|\mathrm{A}|=21-20=1$
$\begin{array}{l}
\therefore \quad \mathrm{A}(\operatorname{adj} \mathrm{A})=\left[\begin{array}{ll}
3 & 4 \\
5 & 7
\end{array}\right]\left[\begin{array}{cc}
7 & -4 \\
-5 & 3
\end{array}\right] \\
=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=1 .\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=|\mathrm{A}| . \mathrm{I}
\end{array}$
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