Question

Let $A=\left\{a_{ij}\right\}$ be a $3\times 3$ matrix, where $a_{ij}=\{\begin{array}{cc}(-1{)}^{j-i}& \text{if}i<j,\\ 2& \text{if}i=j,\\ (-1{)}^{i+j}& \text{if}i>j,\end{array}$
then $det\left(3Adj\right(2A^{-1}\left)\right)$ is equal to

Answer 1

$adj\left(2A^{-1}\right)=|2A^{-1}|(2A^{-1}{)}^{-1}=\frac{8}{|A|}.\frac{1}{2}A=\frac{4A}{|A|}$

So, $|3adj\left(2A^{-1}\right)|=\left|12\frac{A}{|A|}\right|={\left(\frac{12}{|A|}\right)}^3.|A|=\frac{{12}^3}{|A{|}^2}$

$\because A=\left[\begin{array}{ccc}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right]\Rightarrow |A|=4$

Hence, $|3adj\left(2A^{-1}\right)|=\frac{{12}^3}{4^2}=108$