Let $ P $ be a square matrix such that $ P^{2}=I-P $. For $ \alpha, \beta, \gamma, \delta \in \mathbb{N}, $ if $ P^{\alpha}+P^{\beta}=\gamma I-29 P $ and $ P^{\alpha}-P^{\beta}=\delta I-13 P $, then $ \alpha+\beta+\gamma-\delta $ is equal to

JEE MATHEMATICS MCQ: MATRICES

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Analyzing +4 | -1

Let

P

be a square matrix such that

P^{2}=I-P

. For

\alpha, \beta, \gamma, \delta \in \mathbb{N},

P^{\alpha}+P^{\beta}=\gamma I-29 P

and

P^{\alpha}-P^{\beta}=\delta I-13 P

, then

\alpha+\beta+\gamma-\delta

is equal to

View answer and explanation

Correct option: D

The correct answer is D (24) because by repeatedly applying the given relation P² = I - P, we can find expressions for P⁶ and P⁸ in terms of I and P. Then, solving for alpha, beta, gamma, and delta and summing them gives 24.

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

and

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a x+b y+c=0

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

is If the equations

2x + 3y + 1 = 0,3x + y - 2 = 0

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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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2 \times 2

, if

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

, then

|A|

is equal to