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JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Matrices

Let

A = [10 \frac{1}{51} 1]

. If

B = [1 - 1 2 - 1] A [- 1 1 - 2 1]

, then the sum of all the elements of the matrix

\sum_{n = 1}^{50} B^{n}

is equal to

(A)100✓ Correct
(B)50
(C)75
(D)125

Answer: (A)

Solution

Let

C = [1 - 1 2 - 1]

and

D = [- 1 1 - 2 1]

. Then

D C = I

, so

B = C A D

and

B^{n} = C A^{n} D

. By induction,

A^{n} = [10 \frac{n}{51} 1]

. Computing

B^{n} = C A^{n} D

gives

B^{n} = [\frac{n}{51} + 1 - \frac{n}{51} \frac{n}{51} 1 - \frac{n}{51}]

. Summing from

n = 1

to

50

:

n = 1 \sum 50 B^{n} = [50 + \frac{50 \cdot 51}{2 \cdot 51} - \frac{50 \cdot 51}{2 \cdot 51} \frac{50 \cdot 51}{2 \cdot 51} 50 - \frac{50 \cdot 51}{2 \cdot 51}] = [75 - 25 2525] .

Sum of elements =

75 + 25 - 25 + 25 = 100

.

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option

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