Let \(\mathrm{a} \in \mathbf{R}\) and A be a matrix of order \(3 \times 3\) such that \(\operatorname{det}(\mathrm{A})=-4\) and \(\mathrm{A}+\mathrm{I}=\left[\begin{array}{lll}1 & \mathrm{a} & 1 \\ 2 & 1 & 0 \\ \mathrm{a} & 1 & 2\end{array}\right]\), where I is the identity matrix of order \(3 \times 3\). If \(\operatorname{det}((\mathrm{a}+1) \operatorname{adj}((\mathrm{a}-1) \mathrm{A}))\) is \(2^{\mathrm{m}} 3^{\mathrm{n}}, \mathrm{m}, \mathrm{n} \in \{0,1,2, \ldots . .20\}\), then \(\mathrm{m}+\mathrm{n}\) is equal to :

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Matrices

Let

a \in R

and A be a matrix of order

3 \times 3

such that

det (A) = - 4

and

A + I = 12 a a 11 102

, where I is the identity matrix of order

3 \times 3

.

If

det ((a + 1) adj ((a - 1) A))

2^{m} 3^{n}, m, n \in {0, 1, 2, \dots ..20}

, then

m + n

is equal to :

Answer: (D)

Subtracting the identity from

A + I

A = 02 a a 01 101 .

Expanding along the second row,

det A = - 2 (a \cdot 1 - 1 \cdot 1) = - 2 (a - 1)

. Setting this equal to

- 4

gives

a = 3

, so

a + 1 = 4

and

a - 1 = 2

. The quantity sought is

det (4 adj (2 A))

for the

3 \times 3

matrix

2 A

. Using

det (k M) = k^{3} det M

and

det (adj M) = (det M)^{2}

det (4 adj (2 A)) = 4^{3} (det (2 A))^{2} = 64 (2^{3} det A)^{2} = 64 (8 \cdot (- 4))^{2} = 64 \cdot 1024 = 2^{16} .

Writing this as

2^{m} 3^{n} = 2^{16} \cdot 3^{0}

gives

m = 16

and

n = 0

, so

m + n = 16

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option