Let P and Q be 3×3 matrices P ne Q . If {P^3} = {Q^3},{P^2}Q = {Q^2}P then determinant of det

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3 and 4 .Determinants and Matrices

hard

Let $P $ and $Q $ be $3×3$ matrices $P \ne Q$. If ${P^3} = {Q^3},{P^2}Q = {Q^2}P$ then determinant of $\det \left( {{P^2} + {Q^2}} \right)$ is equal to :

A

$-2$

B

$1$

C

$0$

D

$-1$

(AIEEE-2012)

Solution

$(i)$ Two matrices $P$ and $Q$ of order $3 \times 3$ such that

$P \neq Q.$

$(ii)$ $P^{3}=Q^{3}$ and $P^{2} Q=Q^{2} P$

To find The value of determinant of $P^{2}+Q^{2}.$

On subtracting the given equations, we get

${P^{3}-P^{2} Q=Q^{3}-Q^{2} P}$

$\Rightarrow$ ${P^{2}(P-Q)=Q^{2}(Q-P)}$

$\Rightarrow$ $(P-Q)\left(P^{2}+Q^{2}\right)=0$

Now, since $P \neq Q \quad[\text { given }]$

$\Rightarrow \quad P-Q \neq 0 \Rightarrow\left|P^{2}+Q^{2}\right|=0$

$\therefore $ $P^{2}+Q^{2}=0$

Standard 12

Mathematics

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