A) \[0\]
B) \[9\]
C) \[\frac{1}{9}\]
D) \[81\]
Correct Answer: D
Solution :
Given, \[A=\left[ \begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \\ \end{matrix} \right]\] \[{{C}_{11}}=4,\] \[{{C}_{12}}=1,\] \[{{C}_{13}}=-2\] \[{{C}_{21}}=-2,\] \[{{C}_{22}}=4,\] \[{{C}_{23}}=1\] \[{{C}_{31}}=1,\] \[{{C}_{32}}=-2,\] \[{{C}_{33}}=4\] \[\therefore \] \[adj\,(A)=\left[ \begin{matrix} 4 & 1 & -2 \\ -2 & 4 & 1 \\ 1 & -2 & 4 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 4 & -2 & 1 \\ 1 & 4 & -2 \\ -2 & 1 & 4 \\ \end{matrix} \right]\] \[\therefore \] \[|adj\,|A|=4(16+2)+2(4-4)+(1+8)\] \[=72+0+9=81\]
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