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 :
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Official solution
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, Aij=Aji.
For a 3 \[ \times \] 3 symmetric matrix, it looks like this:
\[
abcbdecef
\]
Notice that there are only 6 unique elements we need to fill because of the symmetry:
a in the (1,1) position
b in the (1,2) and (2,1) positions
c in the (1,3) and (3,1) positions
d in the (2,2) position
e in the (2,3) and (3,2) positions
f in the (3,3) position
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.
We have 10 choices for each of the 6 unique elements, so the total number of symmetric matrices can be calculated as:
\[10 \times 10 \times 10 \times 10 \times 10 \times 10 = 10^{6}\]
Thus, the total number of symmetric matrices of order 3 with entries from this set is 106.