Let P and Q be 3 imes 3 matrices P
e Q. If {P^3} = {Q^3} and {P^2}Q = {Q^2}P then determinant of \left( {{P^2} + {Q^2}} ight) is equal to :

Question

Let P and Q be 3 	imes 3 matrices P 
e Q. If {P^3} = {Q^3} and {P^2}Q = {Q^2}P then determinant of \left( {{P^2} + {Q^2}} ight) is equal to :

Accepted Answer

Answer: C. 0 Given {P^3} = {q^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 ight) {P^2}Q = {Q^2}p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 ight) Subtracting (1) and (2), we get {P^3} - {P^2}Q = {Q^3} - {Q^2}P \Rightarrow {P^2}\left( {P - Q} ight) + {Q^2}\left( {P - Q} ight) = 0 \Rightarrow \left( {{P^2} + {Q^2}} ight)\left( {P - Q} ight) = 0 \Rightarrow \left| {{p^2} + {Q^2}} ight| = 0 as P 
e Q

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 :

Answer

Explanation

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