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 :

<p>Sure! A symmetric matrix is a square matrix that is equal to its transpose. For a matrix to be symmetric, the element at row i and column j must be equal to the element at row j and column i. In other words, $A_{ij} = A_{ji}$. </p> <p>For a 3 $\times$ 3 symmetric matrix, it looks like this:</p> <p>$$ \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \\ \end{pmatrix} $$ </p> <p>Notice that there are only 6 unique elements we need to fill because of the symmetry:</p> <ol> <li>$a$ in the (1,1) position</li> <li>$b$ in the (1,2) and (2,1) positions</li> <li>$c$ in the (1,3) and (3,1) positions</li> <li>$d$ in the (2,2) position</li> <li>$e$ in the (2,3) and (3,2) positions</li> <li>$f$ in the (3,3) position</li> </ol> <p>Each of these unique elements can take a value from the set ${0,1,2,3,4,5,6,7,8,9}$, which has 10 elements. </p> <p>We have 10 choices for each of the 6 unique elements, so the total number of symmetric matrices can be calculated as:</p> <p>$10 \times 10 \times 10 \times 10 \times 10 \times 10 = 10^{6}$</p> <p>Thus, the total number of symmetric matrices of order 3 with entries from this set is $10^{6}$.</p>