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

<p>$$P = \left[ {\matrix{ {{a_1}} & {{b_1}} & {{c_1}} \cr {{a_2}} & {{b_2}} & {{c_2}} \cr {{a_3}} & {{b_3}} & {{c_3}} \cr } } \right]$$</p> <p>Given : ${P^T} = aP + (a - 1)I$</p> <p>$$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{b_1}} & {{b_2}} & {{b_3}} \cr {{c_1}} & {{c_2}} & {{c_3}} \cr } } \right] = \left[ {\matrix{ {a{a_1} + a - 1} & {a{b_1}} & {a{c_1}} \cr {a{a_2}} & {a{b_2} + a - 1} & {a{c_2}} \cr {a{a_3}} & {a{b_3}} & {a{c_3} + a - 1} \cr } } \right]$$</p> <p>$$ \Rightarrow {a_1} = a{a_1} + a - 1 \Rightarrow {a_1}(1 - a) = a - 1 \Rightarrow {a_1} = - 1$$</p> <p>Similarly, ${a_1} = {b_2} = {c_3} = - 1$</p> <p>Now, $$\left. \matrix{ {a_2} = a{b_1} \hfill \cr {b_1} = a{a_2} \hfill \cr} \right] \to {a_2} = {a^2}{a_2} \Rightarrow {a_2} = 0 \Rightarrow {b_1} = 0$$</p> <p>${c_1} = a{a_3}$</p> <p>Similarly, all other elements will also be 0</p> <p>${a_2} = {a_3} = {b_1} = {b_3} = {c_1} = {c_2} = 0$</p> <p>$\therefore$ $$P = \left[ {\matrix{ { - 1} & 0 & 0 \cr 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr } } \right]$$</p> <p>$|P| = - 1$</p> <p>$|Adj(P){|_{n \times n}} = |A{|^{(n - 1)}}$</p> <p>$\Rightarrow |Adj(P)| = {( - 1)^2} = 1$</p>