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JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Matrices

Let

A = [1 - 1 24]

. If

A^{- 1} = α I + β A

,

α, β \in R

,

I

is a

2 \times 2

identity matrix, then

4 (α - β)

is equal to :

(A)2
(B)5
(C)4✓ Correct
(D) $\frac{8}{3}$

Answer: (C)

Solution

∣ A ∣ = 1 (4) - 2 (- 1) = 6

A^{- 1} = \frac{1}{6} [41 - 2 1] = [\frac{2}{3} \frac{1}{6} - \frac{1}{3} \frac{1}{6}]

Given

A^{- 1} = α I + β A

:

[\frac{2}{3} \frac{1}{6} - \frac{1}{3} \frac{1}{6}] = [α 0 0 α] + [β - β 2 β 4 β]

Comparing elements:

α + β = \frac{2}{3}

and

2 β = - \frac{1}{3}

So

β = - \frac{1}{6}

and

α = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}

Therefore

4 (α - β) = 4 (\frac{5}{6} + \frac{1}{6}) = 4 (1) = 4

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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