JEE Mathematics 2023 Question with Solution

Given: A symmetric matrix of order $3$ has all entries chosen from the set $\{0,1,2,3,4,5,6,7,8,9\}$.

Find: The number of such symmetric matrices.

A symmetric matrix of order $3$ is of the form

$$A = \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \end{pmatrix}$$

where the condition of symmetry is $a_{ij} = a_{ji}$.

So the independent entries are:

• diagonal entries: $a, d, f$
• off-diagonal entries: $b, c, e$

Hence, the total number of independent entries is $6$.

Each independent entry can be chosen in $10$ ways from $\{0,1,2,3,4,5,6,7,8,9\}$.

Therefore, the total number of symmetric matrices is

$$10 \times 10 \times 10 \times 10 \times 10 \times 10 = 10^6$$

Thus, the number of symmetric matrices is $10^6$. The correct option is B.

Given: The matrix is symmetric and of order $3$.

Find: The number of possible matrices.

For a symmetric matrix, only the entries on and above the diagonal can be chosen freely. In a $3 \times 3$ matrix, that count is

$$\frac{3(3+1)}{2} = 6$$

So there are $6$ independent positions.

Each position has $10$ choices.

Hence,

$$\text{Total number} = 10^6$$

Therefore, the correct option is B.