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If $A=\left[\begin{array}{lll}2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]$, then $|\operatorname{adj} A|$ is equal to
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81
Given, $A=\left[\begin{array}{lll}2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]$
$\mathrm{C}_{11}=4, \quad \mathrm{C}_{12}=1, \quad \mathrm{C}_{13}=-2$
$C_{21}=-2, \quad C_{22}=4, \quad C_{23}=1$
$\mathrm{C}_{31}=1, \quad \mathrm{C}_{32}=-2, \quad \mathrm{C}_{33}=4$
$\therefore \quad \operatorname{adj}(A)=\left[\begin{array}{rrr}4 & 1 & -2 \\ -2 & 4 & 1 \\ 1 & -2 & 4\end{array}\right]^{\mathrm{T}}$
$=\left[\begin{array}{rrr}4 & -2 & 1 \\ 1 & 4 & -2 \\ -2 & 1 & 4\end{array}\right]$
$\therefore \quad|\operatorname{adj} \mathrm{A}|=4(16+2)+2(4-4)+(1+8)$
$=72+0+9=81$
$\mathrm{C}_{11}=4, \quad \mathrm{C}_{12}=1, \quad \mathrm{C}_{13}=-2$
$C_{21}=-2, \quad C_{22}=4, \quad C_{23}=1$
$\mathrm{C}_{31}=1, \quad \mathrm{C}_{32}=-2, \quad \mathrm{C}_{33}=4$
$\therefore \quad \operatorname{adj}(A)=\left[\begin{array}{rrr}4 & 1 & -2 \\ -2 & 4 & 1 \\ 1 & -2 & 4\end{array}\right]^{\mathrm{T}}$
$=\left[\begin{array}{rrr}4 & -2 & 1 \\ 1 & 4 & -2 \\ -2 & 1 & 4\end{array}\right]$
$\therefore \quad|\operatorname{adj} \mathrm{A}|=4(16+2)+2(4-4)+(1+8)$
$=72+0+9=81$
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