The number of symmetric matrices of order 3, with all the entries from the set { 0 , 1 , 2 , 3 , 4 , 5 , 6 ,…

Given that $A=\{0,1,2,3,\dots 8,9\}$

We need to find the number of symmetric matrices.

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

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

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 $10\times 1$.

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

Hence this is the correct option.