Let I be the identity matrix of order $3 \times 3$ and for the matrix $A=\left[\begin{array}{ccc}\lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2\end{array}\right],|A|=-1$. Let $B$ be the inverse of the matrix $\operatorname{adj}\left(\mathrm{A} \operatorname{adj}\left(\mathrm{A}^{2}\right)\right)$. Then $|(\lambda \mathrm{B}+1)|$ is equal to ______

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Matrices

Let I be the identity matrix of order

3 \times 3

and for the matrix

A = λ 47 25 - 1 362, ∣ A ∣ = - 1

. Let

B

be the inverse of the matrix

adj (A adj (A^{2}))

. Then

∣ (λ B + 1) ∣

is equal to ______

Answer: 38

From

∣ A ∣ = - 1

, expanding the determinant gives

λ = 3

. Let

C = A \cdot adj (A^{2})

. Then

A C = A^{2} adj (A^{2}) = ∣ A^{2} ∣ I = I

, so

C = A^{- 1}

. Thus

B^{- 1} = adj (A^{- 1})

, which means

B = (adj (A^{- 1}))^{- 1}

. But

adj (A^{- 1}) = (A^{- 1})^{- 1} /∣ A^{- 1} ∣ \cdot I

... Actually,

B = adj (A)

(from the source). Now

P = 3 B + I = 3 adj (A) + I

. Then

A P = 3 A adj (A) + A = 3∣ A ∣ I + A = A - 3 I

. So

∣ A P ∣ = ∣ A - 3 I ∣

, giving

∣ A ∣∣ P ∣ = ∣ A - 3 I ∣

. Computing

∣ A - 3 I ∣ = - 38

, we get

∣ P ∣ = 38

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Integer answer