Question
Let $$P$$ and $$Q$$ be $$3 \times 3$$ matrices $$P \ne Q.$$ If $${P^3} = {Q^3}$$ and $${P^2}Q = {Q^2}P$$ then determinant of $$\left( {{P^2} + {Q^2}} \right)$$ is equal to:
A.
$$- 2$$
B.
$$1$$
C.
$$0$$
D.
$$- 1$$
Answer :
$$0$$
Solution :
$$\eqalign{
& {\text{Given }}{P^3} = {Q^3}\,\,\,\,.....\left( 1 \right) \cr
& {\text{and }}{P^2}Q = {Q^2}P\,\,\,.....\left( 2 \right) \cr} $$
Subtracting (1) and (2), we get
$$\eqalign{
& {P^3} - {P^2}Q = {Q^3} - {Q^2}P \cr
& \Rightarrow \,\,{P^2}\left( {P - Q} \right) + {Q^2}\left( {P - Q} \right) = 0 \cr
& \Rightarrow \,\,\left( {{P^2} + {Q^2}} \right)\left( {P - Q} \right) = 0 \cr
& \Rightarrow \,\,\left| {{P^2} + {Q^2}} \right| = 0\,\,{\text{as }}P \ne Q \cr} $$