Let $$A = [a_{ij}]$$ and $$B = [b_{ij}]$$ be two $$3 \times 3$$ real matrices such that $$b_{ij} = (3)^{(i+j-2)} a_{ij}$$, where $$i, j = 1, 2, 3$$. If the determinant of B is 81, then determinant of A is

We are told that the two real matrices $$A=[a_{ij}]$$ and $$B=[b_{ij}]$$ of order $$3 \times 3$$ satisfy the relation

$$b_{ij}=3^{\,i+j-2}\,a_{ij},\qquad i,j=1,2,3.$$

First we recognise that the factor $$3^{\,i+j-2}$$ attached to $$a_{ij}$$ can be split into a product of a row-dependent factor and a column-dependent factor:

$$3^{\,i+j-2}=3^{\,i-1}\,3^{\,j-1}.$$

Now let us introduce two diagonal matrices

$$L=\operatorname{diag}\bigl(3^{\,0},3^{\,1},3^{\,2}\bigr),\qquad R=\operatorname{diag}\bigl(3^{\,0},3^{\,1},3^{\,2}\bigr).$$

The entry in the $$i^{\text{th}}$$ row and $$j^{\text{th}}$$ column of the product $$LAR$$ is

$$\bigl(LAR\bigr)_{ij}=L_{ii}\,A_{ij}\,R_{jj}=3^{\,i-1}\,a_{ij}\,3^{\,j-1}=3^{\,i+j-2}\,a_{ij}=b_{ij}.$$

Thus we have the matrix equality

$$B=LAR.$$

We now take determinants on both sides. Using the basic property “determinant of a product equals the product of determinants” we obtain

$$\det(B)=\det(L)\,\det(A)\,\det(R).$$

Because both $$L$$ and $$R$$ are diagonal, their determinants are simply the products of their diagonal entries:

$$\det(L)=3^{\,0}\cdot3^{\,1}\cdot3^{\,2}=1\cdot3\cdot9=27,$$

$$\det(R)=3^{\,0}\cdot3^{\,1}\cdot3^{\,2}=27.$$

Substituting these values we get

$$\det(B)=27\;\times\;\det(A)\;\times\;27 = 729\,\det(A).$$

We are given that $$\det(B)=81$$, so

$$81=729\,\det(A).$$

Dividing both sides by $$729$$ gives

$$\det(A)=\frac{81}{729}=\frac{1}{9}.$$

Hence, the correct answer is Option D.