The number of symmetric matrices of order 3, with all the entries from the set 0,1,2,3,4,5,6,7,8,9 is

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

We need to find the number of symmetric matrices.

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

$$A=[\begin{matrix}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{matrix}]$$A=[a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​]

Now ${a}_{11},{a}_{22},{a}_{33}$a11​,a22​,a33​ can have any digit from $0$0 to $9$9.

Number of ways this can be done is $10\times 10\times 10={10}^{3}$10×10×10=103

Now ${a}_{12}={a}_{21}$a12​=a21​. So number of ways this can be done is $10\times 1$10×1.

Similarly ${a}_{13}={a}_{31}$a13​=a31​, So number of ways this can be done is $10\times 1$10×1.

Also, ${a}_{23}={a}_{32}$a23​=a32​. So number of ways this can be done is $10\times 1$10×1.

$\therefore$∴ Total symmetric matrix$={10}^{3}\times 10\times 10\times 10={10}^{6}$=103×10×10×10=106

Hence this is the correct option.