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JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Matrices

If

A = [1 λ 510]

,

A^{- 1} = α A + β I

and

α + β = - 2

, then

4 α^{2} + β^{2} + λ^{2}

is equal to:

(A)12
(B)10
(C)19
(D)14✓ Correct

Answer: (D)

Solution

By the Cayley--Hamilton theorem

A^{2} - (tr A) A + (det A) I = 0

, with

tr A = 11

and

det A = 10 - 5 λ

. Multiplying by

A^{- 1}

,

A^{- 1} = \frac{1}{10 - 5 λ} (11 I - A) = \frac{- 1}{10 - 5 λ} A + \frac{11}{10 - 5 λ} I .

Comparison with

A^{- 1} = α A + β I

gives

α = \frac{- 1}{10 - 5 λ}

and

β = \frac{11}{10 - 5 λ}

, so

α + β = \frac{10}{10 - 5 λ} = - 2 \Rightarrow 10 - 5 λ = - 5 \Rightarrow λ = 3.

Then

α = \frac{1}{5}

and

β = - \frac{11}{5}

, giving

4 α^{2} + β^{2} + λ^{2} = \frac{4}{25} + \frac{121}{25} + 9 = 5 + 9 = 14.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option

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