If P is a 33 real matrix such that P^T=aP+(a-1)I, where a1, then

Question

If $P$ is a $3	imes3$ real matrix such that $P^T=aP+(a-1)I$, where $a>1$, then

Accepted Answer

Taking the transpose of $P^T=aP+(a-1)I$: $P=aP^T+(a-1)I$. Subtracting: $P^T-P=a(P-P^T)\Rightarrow (P^T-P)(1+a)=0$. Since $a>1$, $a
e-1$, so $P^T=P$ ($P$ is symmetric).

Then $P=aP+(a-1)I\Rightarrow(1-a)P=(a-1)I\Rightarrow P=-I$.

So $|P|=(-1)^3=-1$, and $|\operatorname{adj}P|=|P|^{3-1}=|P|^2=1$.

Solutions · 1