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Question
If \[A = \begin{bmatrix}1 & 2 & - 1 \\ - 1 & 1 & 2 \\ 2 & - 1 & 1\end{bmatrix}\] , then ded (adj (adj A)) is __________ .
Options
144
143
142
14
Solution
144
Given:
\[A = \begin{bmatrix} 1 & 2 & - 1\\ - 1 & 1 & 2\\ 2 & - 1 & 1 \end{bmatrix}\]
\[ \therefore \left| A \right| = \begin{vmatrix} 1 & 2 & - 1\\ - 1 & 1 & 2\\ 2 & - 1 & 1 \end{vmatrix} = 1\left( 1 + 2 \right) - 2\left( - 1 - 4 \right) - 1\left( 1 - 2 \right) = 3 + 10 + 1 = 14\]
We have
\[\left| adj\left( adj A \right) \right| = \left| A \right|^{( n - 1)^2} \]
\[ \Rightarrow \left| adj\left( adj A \right) \right| = \left( 14 \right)^{2^2} = {14}^4 \]
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