A) \[1\]
B) \[-1\]
C) \[\frac{1}{3}\]
D) \[3\]
Correct Answer: A
Solution :
Given, \[A=\left( \begin{matrix} 1 & 2 \\ 2 & 1 \\ \end{matrix} \right)\] Then, \[adj\,(A)=\left( \begin{matrix} 1 & -2 \\ -2 & 1 \\ \end{matrix} \right)\] and \[adj\,\,\,(adj\,A)=\left( \begin{matrix} 1 & 2 \\ 2 & 1 \\ \end{matrix} \right)\] Now, \[A\,(adj\,(adj(A))=\left( \begin{matrix} 1 & 2 \\ 2 & 1 \\ \end{matrix} \right)\,\left( \begin{matrix} 1 & 2 \\ 2 & 1 \\ \end{matrix} \right)\] \[=\left( \begin{matrix} 1+4 & 2+2 \\ 2+2 & 4+1 \\ \end{matrix} \right)\] \[=\left( \begin{matrix} 5 & 4 \\ 4 & 5 \\ \end{matrix} \right)\] \[\Rightarrow \] \[\frac{1}{3}A(adj\,(adj\,A))=\left( \begin{matrix} 5/3 & 4/3 \\ 4/3 & 5/3 \\ \end{matrix} \right)\] Determinant of \[\frac{1}{3}\,A(adj\,(adj\,A))=\left| \begin{matrix} 5/3 & 4/3 \\ 4/3 & 5/3 \\ \end{matrix} \right|\] \[=\frac{5}{3}\times \frac{5}{3}-\frac{4}{3}\times \frac{4}{3}=\frac{25}{9}-\frac{16}{9}\] \[=\frac{9}{9}=1\]
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