The number of symmetric matrices of order 3, with all the entries...

Given that $A = { 0, 1, 2, 3, \ldots 8, 9 }$

We need to find the number of symmetric matrices.

We know that a matrix is symmetric if ${A}^{τ} = A$.

A = \left[\begin{cases}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{cases}\right]

Now ${a}_{11}, {a}_{22}, {a}_{33}$ can have any digit from $0$ to $9$.

Number of ways this can be done is $10 \times 10 \times 10 = {10}^{3}$

Now ${a}_{12} = {a}_{21}$. So number of ways this can be done is $10 \times 1$.

Similarly ${a}_{13} = {a}_{31}$, So number of ways this can be done is $10 \times 1$.

Also, ${a}_{23} = {a}_{32}$. So number of ways this can be done is $10 \times 1$.

$∴$ Total symmetric matrix$= {10}^{3} \times 10 \times 10 \times 10 = {10}^{6}$

Hence this is the correct option.