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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
