Explanation

A matrix is symmetric if it is equal to its transpose. That means:

$$A = A^T$$

For this 3×3 matrix:

$$A = begin{bmatrix} 5 & x & 3 \ x & 7 & 2 \ 3 & 2 & 9 \ end{bmatrix}$$

We compare element aij​ with $a_{ji}$ The (1,2) element is $x$, and the (2,1) element is also $x$, so they're already equal.
Now, look at element (1,3), which is 3, and element (3,1), which is also 3 — so they match.
Now check (2,3) = 2 and (3,2) = 2 — they also match.

So, the matrix is symmetric if and only if:

• $x=x$ (always true)

• All off-diagonal symmetric pairs match

Given that, the matrix is symmetric already, but if in the original problem the matrix looked like this:

$$begin{bmatrix} 5 & x & 3 \ 11 & 7 & 2 \ 3 & 2 & 9 \ end{bmatrix}$$

Then, to make it symmetric:

• The element at (1,2) must equal the element at (2,1)

• So: $$x=11