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Verify $A(\operatorname{adj} A)=(\operatorname{adj} A) \cdot A=|A| I$
$\left[\begin{array}{cc}2 & 3 \\ -4 & 6\end{array}\right]=\mathbf{A}(\mathbf{s a y})$
$\left[\begin{array}{cc}2 & 3 \\ -4 & 6\end{array}\right]=\mathbf{A}(\mathbf{s a y})$
Solution:
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Verified Answer
$\begin{aligned}
&|\mathrm{A}|=24 \\
&\mathrm{C}_{11}=(-1)^{1+1}(-6)=6 ; \mathrm{C}_{12}=(-1)^{1+2}(-4)=4
\end{aligned}$
$C_{21}=(-1)^{2+1}(3)=-3 ; C_{22}^{12}=(-1)^{2+2}(2)=-2$
$\therefore \quad \operatorname{Adj} A=\left[\begin{array}{cc}6 & -3 \\ 4 & 2\end{array}\right]$ $\operatorname{AAdj} A=\left[\begin{array}{cc}2 & 3 \\ -4 & 6\end{array}\right]\left[\begin{array}{cc}6 & -3 \\ 4 & 2\end{array}\right]=\left[\begin{array}{cc}24 & 0 \\ 0 & 24\end{array}\right]=24 \mathrm{I}$
Hence $A(\operatorname{Adj} A)=(\operatorname{Adj} A) \cdot A=24 I=|A| I$
&|\mathrm{A}|=24 \\
&\mathrm{C}_{11}=(-1)^{1+1}(-6)=6 ; \mathrm{C}_{12}=(-1)^{1+2}(-4)=4
\end{aligned}$
$C_{21}=(-1)^{2+1}(3)=-3 ; C_{22}^{12}=(-1)^{2+2}(2)=-2$
$\therefore \quad \operatorname{Adj} A=\left[\begin{array}{cc}6 & -3 \\ 4 & 2\end{array}\right]$ $\operatorname{AAdj} A=\left[\begin{array}{cc}2 & 3 \\ -4 & 6\end{array}\right]\left[\begin{array}{cc}6 & -3 \\ 4 & 2\end{array}\right]=\left[\begin{array}{cc}24 & 0 \\ 0 & 24\end{array}\right]=24 \mathrm{I}$
Hence $A(\operatorname{Adj} A)=(\operatorname{Adj} A) \cdot A=24 I=|A| I$
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