Let A be a $2 \times 2$ matrix with real entries such that $\mathrm{A}^{\prime}=\alpha \mathrm{A}+\mathrm{I}$, where $\alpha \in \mathbb{R}-\{-1,1\}$. If $\operatorname{det}(A^{2}-A)=4$, then the sum of all possible values of $\alpha$ is equal to

Matrices — JEE Main · 2023 · 11 Apr 2023 · Shift 1 Question with Solution

Let A be a

2 \times 2

matrix with real entries such that

A^{'} = α A + I

, where

α \in R - {- 1, 1}

. If

det (A^{2} - A) = 4

, then the sum of all possible values of

α

is equal to

Answer: (C)

From

A^{T} = α A + I

, taking transpose:

A = α A^{T} + I = α (α A + I) + I = α^{2} A + (α + 1) I

. Thus

A (1 - α^{2}) = (α + 1) I

, giving

A = \frac{I}{1 - α}

(since

α \neq = \pm 1

). Then

∣ A ∣ = \frac{1}{( 1 - α ) ^{2}}

and

A - I = \frac{α}{1 - α} I

, so

∣ A - I ∣ = (\frac{α}{1 - α})^{2}

. From

∣ A^{2} - A ∣ = ∣ A ∣∣ A - I ∣ = 4

\frac{1}{( 1 - α ) ^{2}} \cdot \frac{α ^{2}}{( 1 - α ) ^{2}} = 4

, giving

\frac{α ^{2}}{( 1 - α ) ^{4}} = 4

, so

\frac{α}{( 1 - α ) ^{2}} = \pm 2

. Case 1:

2 (1 - α)^{2} = α \Rightarrow 2 α^{2} - 5 α + 2 = 0

, giving

α = 2

\frac{1}{2}

. Case 2:

2 (1 - α)^{2} = - α \Rightarrow 2 α^{2} - 3 α + 2 = 0

, which has no real solutions. Sum of all possible values:

2 + \frac{1}{2} = \frac{5}{2}

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Single correct option