Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]$ and $\mathrm{B}=\left[\mathrm{b}_{\mathrm{ij}}\right]$ be two $3 \times 3$ real matrices such that $\mathrm{b}_{\mathrm{ij}}=(3)^{(\mathrm{i}+\mathrm{j}-2)} \mathrm{a}_{\mathrm{ji}}$, where $i, j=1,2,3$. If the determinant of $B$ is 81, then the determinant of $A$ is:

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Determinants

Let

A = [a_{ij}]

and

B = [b_{ij}]

be two

3 \times 3

real matrices such that

b_{ij} = (3)^{(i + j - 2)} a_{ji}

, where

i, j = 1, 2, 3

. If the determinant of

B

is 81, then the determinant of

A

is:

Answer: (B)

The rule

b_{ij} = 3^{i + j - 2} a_{j i}

builds

B

from the transpose of

A

with each entry multiplied by a power of

3

. Pulling the common factors

3^{0}, 3^{1}, 3^{2}

out of the three rows and again out of the three columns multiplies the determinant by

3^{0 + 1 + 2} \cdot 3^{0 + 1 + 2} = 27 \times 27 = 729

, so

∣ B ∣ = 729 ∣ A^{T} ∣ = 729 ∣ A ∣

. With

∣ B ∣ = 81

∣ A ∣ = \frac{81}{729} = \frac{1}{9}

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option