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JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Matrices

Let the matrix

A = 001100010

and the matrix

B_{0} = A^{49} + 2 A^{98}

. If

B_{n} = Adj (B_{n - 1})

for all

n \geq 1

, then

det (B_{4})

is equal to :

(A) $3^{28}$
(B) $3^{30}$
(C) $3^{32}$ ✓ Correct
(D) $3^{36}$

Answer: (C)

Solution

First, compute

A^{2}

and

A^{3}

to find that

A^{3} = I

. Therefore,

A^{49} = A^{48} \cdot A = (A^{3})^{16} \cdot A = A

and

A^{98} = A^{96} \cdot A^{2} = (A^{3})^{32} \cdot A^{2} = A^{2}

.

Since

A^{2} = I

(after row operations), we have

B_{0} = A + 2 I = 201120012

.

Compute

det (B_{0}) = 2 (4 - 0) - 1 (0 - 1) = 8 + 1 = 9

.

For an

n \times n

matrix

M

,

det (Adj (M)) = (det M)^{n - 1}

. Applying this recursively:

det (B_{4}) = det (Adj (Adj (Adj (Adj (B_{0}))))) = (det B_{0})^{(n - 1)^{4}} = 9^{16} = (3^{2})^{16} = 3^{32}

.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option

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