Let $P$ and $Q$ be $3 \times 3$ matrices with $P \neq Q.$ If $P^3=Q^3$ and $P^2 Q=Q^2 P,$ the determinant of $\left(P^2+Q^2\right)$ is equal to

JEE MATHEMATICS MCQ: MATRICES

1 Easy

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Analyzing +4 | -1

Let

P

and

Q

3 \times 3

matrices with

P \neq Q.

P^3=Q^3

and

P^2 Q=Q^2 P,

the determinant of

\left(P^2+Q^2\right)

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct. By subtracting the given equations and factoring, it's shown that the determinant of (P² + Q²) must be 0 since P is not equal to Q.

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{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

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\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

, is If

\omega

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\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

is equal to For any matrix

A

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, if

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

, then

|A|

is equal to