JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Matrices

Let

A = 100 - 1 10 0 - 1 1

and

B = 7 A^{20} - 20 A^{7} + 2 I

, where

I

is an identity matrix of order

3 \times 3

. If

B = [b_{ij}]

, then

b_{13}

is equal to........

Answer: 910

Solution

Powers of

A

keep the same shape,

A^{n} = 100 - n 10^{n} C_{2} - n 1

which holds for

n = 1

and carries to

n + 1

on one more multiplication by

A

, since

^{n} C_{2} + n =^{n + 1} C_{2}

. The identity matrix carries a zero in the first row and third column, so the term

2

times the identity leaves

b_{13}

untouched, and

b_{13} = 7^{20} C_{2} - 20^{7} C_{2} = 7 (190) - 20 (21) = 1330 - 420 = 910

giving

b_{13} = 910

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer