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JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Matrices

Let

A = {a_{ij}}

be a

3 \times 3

matrix, where

a_{ij} = ⎩ ⎨ ⎧ (- 1)^{j - i} 2 (- 1)^{i + j} if if if i < j, i = j, i > j,

then

det (3 Adj (2 A^{- 1}))

is equal to

Answer: 108

Solution

The rule gives

a_{12} = (- 1)^{1} = - 1

,

a_{13} = (- 1)^{2} = 1

,

a_{23} = (- 1)^{1} = - 1

above the diagonal,

a_{21} = (- 1)^{3} = - 1

,

a_{31} = (- 1)^{4} = 1

,

a_{32} = (- 1)^{5} = - 1

below it, and

2

on the diagonal:

A = 2 - 1 1 - 1 2 - 1 1 - 1 2, det A = 2 (4 - 1) + 1 (- 2 + 1) + 1 (1 - 2) = 6 - 1 - 1 = 4

For a

3 \times 3

matrix

B

,

det (k B) = k^{3} det B

and

det (Adj B) = (det B)^{2}

. Hence

det (2 A^{- 1}) = \frac{2 ^{3}}{det A} = \frac{8}{4} = 2, det (Adj (2 A^{- 1})) = 2^{2} = 4

and

det (3 Adj (2 A^{- 1})) = 3^{3} \times 4 = 108

.

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Integer answer

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