"Tricks" for solving the determinant of a matrix

Question

I've got an exam coming up, and I'm worried that (as I usually tend to do) I will mess up with some minor algebraic calculation in a large problem like finding the determinant of a 4x4 or larger matrix. I'm likely to be asked to use Laplace Expansion, and I'd like to know if there are any other quick and dirty ways I can try to make sure my answer was correct. Considering I'm not alowed to use a calculator, is there some other non-standard method I can use to try to verify a correct solution?

I know how to find the determinant no problem, but I'm very prone to minor mistakes and I don't want to have to spend too much time reading over every problem in fine detail to make sure I didn't omit a negative sign somewhere.

Answer 1

If you have time, you could always do the calculation twice, once with the top row as a starting point and one (say) with the bottom row. For example: $$\begin{vmatrix} a & b & c\\d & e & f\\g & h & i \end{vmatrix}=$$ $$a(ei - fh) -b(di -fg) + c(dh - eg)$$ Or: $$g(bf - ce) - h(af -cd) +i(ae -bd)$$

Of course these give the same result, just with a different order of calculations. If you calculate both by hand and get different results, you know you have an error.

The method suggested bit Git Gud in the comments can also be used, i.e. add scalar multiples of the different rows to each other to get a triangular matrix. A worked example may be found here.

Answer 2

If the matrix is structured so that a certain row or column has a lot of zeros in it then you should be sure to take advantage of this. Expanding along such a row/column will reduce the number of calculations you must perform, which will save time and simultaneously reduce the risk of error.

Answer 3

There's not a standard way to compute determinant avoiding those errors, but you can use some tricks before Laplace Expansion to simplify your matrix. Anyway, the tricks you can use depends on your matrix. For example, if it has two similar rows, you can reduce by rows to obtain a lot of zeros which makes simpler your computation. Instead, if using Gaussian elimination you can get quickly a triangular matrix, you can use it and then compute determinant taking the product of diagonal entries.

Answer 4

You might consider Pivotal Condensation.

PC reduces an $n\times n$ determinant to an $(n-1)\times(n-1)$ determinant whose entries happen to be $2\times 2$ determinants. Simply iterate until your determinant gets to reasonable size. (You can/should stop at $3\times 3$, at which point it's easy enough to compute the final result manually.)

Web-searching "pivotal condensation" will give you lots of explanations and examples. This document even includes a simple proof. This lengthy YouTube video purports to talk through the process, but I didn't view the whole thing.

Pivotal condensation can be extremely tedious; it may-or-may-not be time-effective in an exam situation. However, at every stage of the process, you compute mere $2\times 2$ determinants, which are super-easy and super-quick. So there's a trade-off. (Of course, there's another trade-off: with so many to compute, the danger of stupid arithmetic mistakes accumulates.)

I've actually found pivotal condensation helpful with certain symbolic matrices (in particular, ones with trigonometric terms), since it gives me a systematic way to (attempt to) combine elements. In that regard, it's one of my favorite "tricks".