If P is a 3 times 3 real matrix such that P^{ | vedclass

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(D) Given $P^{T} = aP + (a - 1)I$. Taking transpose on both sides,we get $(P^{T})^{T} = (aP + (a - 1)I)^{T}$. $P = aP^{T} + (a - 1)I$. Substitute $P^{

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$|\operatorname{Adj} P| = 1$

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(D) Given $P^{T} = aP + (a - 1)I$. Taking transpose on both sides,we get $(P^{T})^{T} = (aP + (a - 1)I)^{T}$. $P = aP^{T} + (a - 1)I$. Substitute $P^{T} = aP + (a - 1)I$ into the equation: $P = a(aP + (a - 1)I) + (a - 1)I$. $P = a^{2}P + a(a - 1)I + (a - 1)I$. $P = a^{2}P + (a^{2} - a + a - 1)I$. $P = a^{2}P + (a^{2} - 1)I$. $(1 - a^{2})P = (a^{2} - 1)I$. Since $a > 1$,$a^{2} - 1 
eq 0$,so $-(a^{2} - 1)P = (a^{2} - 1)I$. $P = -I$. Now,$|P| = |-I| = (-1)^{3} |I| = -1$. We know that $|\operatorname{Adj} P| = |P|^{n-1}$,where $n = 3$. $|\operatorname{Adj} P| = (-1)^{3-1} = (-1)^{2} = 1$.

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

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