A) \[{{14}^{4}}\]
B) \[{{14}^{3}}\]
C) \[{{14}^{2}}\]
D) \[14\]
Correct Answer: A
Solution :
We have,\[A=\left[ \begin{matrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[|A|=1(1+2)-2(-1-4)-1(1-2)\] \[=3+10+1=14\] We know that, for a square matrix of order n, \[adj(adj\,\,A)=|A{{|}^{n-2}}A,\]if\[|A|\ne 0\] \[\Rightarrow \] \[\det (adj(adj\,\,A))=||A{{|}^{n-2}}A|\] \[\Rightarrow \] \[\det (adj(adj\,\,A))={{(|A{{|}^{n-2}})}^{n}}|A|\] \[\Rightarrow \] \[\det (adj(adj\,\,A))=|A{{|}^{{{n}^{2}}-2n+1}}\] Here, \[n=3\]and\[|A|=14\] \[\therefore \] \[\det (adj\,\,(adjA))={{(14)}^{{{3}^{2}}-2\times 3+1}}={{14}^{4}}\]
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