$B = A^T A$ can only be one of the following. Which one? I know that the answer is (b) but I have no idea on how to arrive there. What would be a good approach to arrive at the answer?

 A: The determinant of $A^TA$ is the product $\det A\times\det A^T=(\det A)^2$, so must be $>0$.
In the five options there is only one satisfying this criterion.

Hope this helps
A: I would check it in this order:


*

*$B$ needs to be symmetric, which rules out (c). 

*$B$ needs to be nonsingular (as $A$ was nonsingular) which rules out (d).

*The diagonal of $B$ carries norms and hence cannot contain negative values (which rules out (e)).

*$B$ needs to be positive definite which rules out (a) (negative determinant).


In the particular example you show, the determinant condition actually rules out all options except (b), as awllower mentioned. This does not always need to be the case. For example, $B = -I_2$ has determinant +1 and yet is not positive definite.
A: This can be deduced as follows. Note that the product of A with it's transpose is symmetric. Thus one can eliminate C. Also if A is nonsingular, then B is also non singular. Thus, the determinant cannot be zero which eliminates D. Finally remember the determinant of transpose(A) is equal to that of A. This implies that the determinant of B is positive since it is equal to (det(A))^2. Therefore B is the correct answer. Hope this helps.
