If P is a 3 × 3 real matrix such that P T = a P + ( a - 1 ) I , where a > 1 , then

Given,

$P^T=aP+(a-1)\mathrm{I}......\left(1\right)$

Taking transpose on both sides we get,

${\left(P^T\right)}^T={\left(\mathrm{aP}+(\mathrm{a}-1)\mathrm{I}\right)}^T$

$\Rightarrow P=a{P}^{\mathrm{T}}+(a-1)\mathrm{I}.......\left(2\right)$

Subtract equation (1) and (2),

$\Rightarrow P^T-P=a\left(P-P^T\right)$

$\Rightarrow P=P^{\mathit{T}}$ as $a\ne -1$

From equation (1),

$P=aP+(a-1)I$

$\Rightarrow P=-I$

Taking adjoint on both sides,

$\Rightarrow \mathrm{Adj} P=\mathrm{Adj}(-I)$

$\Rightarrow \left|\mathrm{Adj} P\right|=\left|\mathrm{Adj} (-I)\right|={\left|-I\right|}^2$

$\Rightarrow |AdjP|=1$