Show that B is a nonsingular matrix (not that obvious). How would you proceed if you were asked in an interview to show that B is a nonsingular matrix (in an elegant way)?
$$B= \begin{pmatrix} 1& 1.25& −0.50& 0.15\\ 0.15& 2& 1.25& −1.50\\ −0.45& 0.25& 3& 1.25\\ 0.25& −0.15& 0.25& 4\\ \end{pmatrix}$$
In my opinion, taking the time to compute the determinant of this $4\times4$ matrix during the interview would not be appreciated by the interviewer.
 A: $B^T$ is non-singular because it is a strict diagonally dominant matrix. So $B$ is non-singular as well.
A: Another approach:
Multiply by a constant to make it an integer matrix: in this case, consider the matrix $20B$, which is an integer matrix. Show that its determinant is non-zero modulo some small prime, eg: 2 or 3.
A: Something that will always work with low complexity and not dependent on which matrix:


*

*Take a random vector

*Loop over columns / rows: 


*

*Remove projection with current row/col



If after three (or $n-1$) removed projections is not parallel to last vector ( up to precision ) then there is a non-zero sized null-space.
I'm very sure that the chance of accidentally hitting a subspace in the value space is abysmally small.
A: Another possibility: 
Use Gaussian elimination on $B$ to yield an upper-triangular matrix. Nonzeros on the diagonal of the resulting upper-triangular matrix means that $B$ is nonsingular.
(OP asked for an elegant way to do this; however, "elegant" was not defined. IMHO, Gaussian elimination is way more elegant than computing the determinant.)
