Matrices · Question 15

JEE Main 2025 (Online) 2nd April Morning ShiftMCQMedium📈 Topic in 10% papers+4−1

Let A = bmatrix & -1 \ 6 & bmatrix, > 0, such that (A) = 0 and + = 1. If I denotes 2 2 identity matrix, then the matrix (I + A)8 is :

bmatrix 257 & -64 \ 514 & -127 bmatrix
bmatrix 766 & -255 \ 1530 & -509 bmatrix
bmatrix 1025 & -511 \ 2024 & -1024 bmatrix
bmatrix 4 & -1 \ 6 & -1 bmatrix

Answer

(2) bmatrix 766 & -255 \ 1530 & -509 bmatrix

Solution

(A) = + 6 = 0 and = 1 -

(1 - ) + 6 = 0 2 - - 6 = 0

( - 3)( + 2) = 0 = 3 ( > 0), = -2

A = bmatrix 3 & -1 \ 6 & -2 bmatrix

A2 = bmatrix 9 - 6 & -3 + 2 \ 18 - 12 & -6 + 4 bmatrix = bmatrix 3 & -1 \ 6 & -2 bmatrix = A

Ak = A for every k 1 and IA = AI, so by the binomial theorem:

(I + A)8 = I + ( 81 + 82 + + 88 )A

= I + (28 - 1)A = I + 255A

= bmatrix 1 + 765 & -255 \ 1530 & 1 - 510 bmatrix = bmatrix 766 & -255 \ 1530 & -509 bmatrix

Answer: (2)

JEE Main 2025 (Online) 2nd April Morning Shift

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Let A = bmatrix 1 2 & -2 \ 0 & 1 bmatrix and P = bmatrix & - \ & bmatrix, > 0. If B = PAP , C = P B10P and the sum of the diagonal elements of C is mn, where (m, n) = 1, then m + n is :

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JEE Main 2025 (Online) 28th January Evening Shift

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