If a matrix has 13 elements, then the possible dimensions (orders) of the matrix are
(a) $13\times 1\text{ or }1\times 13$
(b) $1\times 26\text{ or 26}\times \text{1}$
(c) $2\times 13\text{ or 13}\times \text{2}$
(d) $13\times 13$

Hint: The point to focus is that 13 is a prime number and we know that the number of elements in a matrix is the product of its dimensions, i.e., the number of rows multiplied by the number of columns. Use the constraint that the number of elements is a non-negative integer.

Complete step-by-step answer: Let us start the solution to the above question by using the fact that 13 is a prime number, i.e., it is only divisible by itself and 1.
Now, we know that the dimensions of a matrix are its number of rows multiplied by the number of columns and the number of elements is the result of the multiplication. So, we know that the number of elements, number of rows and number of columns are integers.
Now, we let the dimension of the matrix be $x\times y$ . So, according to the above constraints, we can say that x, y are non-negative integers, and $xy=13$ .
Now, there are only two sets (x,y) satisfying the condition mentioned above: (13,1) and (1,13). So, the dimensions of the matrix can be $13\times 1\text{ or }1\times 13$ .

So, the correct answer is “Option A”.

Note: If you want you can solve the above question using the method of option elimination. Just use the point that the product of the dimensions is equal to the number of elements and check for which option the product of the dimensions is coming to be 13, only option (a) will satisfy the condition which is our answer.