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 :
JEE · Math · previous-year question
- A.$$10^{9}$$
- B.$$9^{10}$$
- C.$$10^{6}$$correct
- D.$$6^{10}$$
Answer
C. $$10^{6}$$
Explanation
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, $$A_{ij} = A_{ji}$$. For a 3 $$ \times $$ 3 symmetric matrix, it looks like this: $$ \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \\ \end{pmatrix} $$ 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 $$10^{6}$$.
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