Let A be a square matrix of order 3 such that $\operatorname{det}(\mathrm{A})=-2$ and $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 \mathrm{A})))=2^{\mathrm{m}+\mathrm{n}} \cdot 3^{\mathrm{mn}}$, where $\mathrm{m}>\mathrm{n}$. Then $4 \mathrm{m}+2 \mathrm{n}$ is equal to ____.

Matrices — JEE Main · 2025 · 22 Jan 2025 · Shift 1 Question with Solution

Let A be a square matrix of order 3 such that

det (A) = - 2

and

det (3 adj (- 6 adj (3 A))) = 2^{m + n} \cdot 3^{mn}

, where

m > n

. Then

4 m + 2 n

is equal to ____.

Answer: 34

A

3 \times 3

with

det (A) = - 2

. Use

det (k M) = k^{n} det (M)

and

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

for an

n \times n

matrix (here

n = 3

, so the adjoint exponent is 2).

det (3 A) = 3^{3} det (A) = 27 (- 2) = - 2 \cdot 3^{3} .

det (adj (3 A)) = [det (3 A)]^{2} = (- 2 \cdot 3^{3})^{2} = 2^{2} \cdot 3^{6} .

Let

M = - 6 adj (3 A)

det (M) = (- 6)^{3} det (adj (3 A)) = - (2^{3} \cdot 3^{3}) (2^{2} \cdot 3^{6}) = - (2^{5} \cdot 3^{9}) .

det (adj (M)) = det (M)^{2} = (2^{5} \cdot 3^{9})^{2} = 2^{10} \cdot 3^{18} .

Finally, for

N = 3 adj (M)

det (N) = 3^{3} det (adj (M)) = 3^{3} \cdot 2^{10} \cdot 3^{18} = 2^{10} \cdot 3^{21} .

Comparing with

2^{m + n} \cdot 3^{mn}

m + n = 10

and

mn = 21

. These give

m, n

as roots of

t^{2} - 10 t + 21 = 0

, i.e.

t = 3

t = 7

. Since

m > n

m = 7, n = 3

4 m + 2 n = 28 + 6 = 34.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Integer answer