If P is a $3 \times 3$ real matrix such that $\mathrm{P}^{\mathrm{T}}=\mathrm{aP}+(\mathrm{a}-1) \mathrm{I}$, where $\mathrm{a}>1$, then

(D) $|\text{Adj } \mathrm{P}|=1$

(B) $|\text{Adj } \mathrm{P}|>1$

(C) $|\operatorname{Adj} P|=\frac{1}{2}$

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Matrices

If P is a

3 \times 3

real matrix such that

P^{T} = aP + (a - 1) I

, where

a > 1

, then

Answer: (D)

Transposing the relation

P^{T} = a P + (a - 1) I

gives

P = a P^{T} + (a - 1) I

. Substituting the original relation for

P^{T}

into this,

P = a [a P + (a - 1) I] + (a - 1) I = a^{2} P + (a - 1) (a + 1) I

, so that

(1 - a^{2}) P = (a^{2} - 1) I

. For

a > 1

the factor

1 - a^{2} \neq = 0

, hence

P = - I

. Then

∣ P ∣ = (- 1)^{3} = - 1

, and for a

3 \times 3

matrix

∣ Adj P ∣ = ∣ P ∣^{3 - 1} = ∣ P ∣^{2} = 1

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option