15.4 Evaluating Determinants

Well, from the first two facts alone, we can compute the value of any determinant, and hence the area or volume or whatever of any parallel sided figure.

How?

Well, we can add multiples of rows to one another to get rid of off diagonal elements. When we are done, we can deduce the value of the determinant to be the product of the diagonal elements.

Actually, we need only make the elements on one side of the diagonal into $$0$$, and take the product of the diagonal elements. Getting rid of the others is sometimes a nice thing to do, but will not affect the diagonal elements at all.

Let's evaluate the determinants of the following arrays:

$\begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix}, \begin{pmatrix} 2 & 2 \\ 5 & 7 \end{pmatrix}$

If we subtract $$5$$ times the first row from the second in the first matrix we get $$(0, -3)$$ for the second row, so the determinant is $$-3$$. In the second array we subtract $$\frac{5}{2}$$ times the first row from the second and get $$(0, 2)$$ as new second row. The determinant of the second matrix is therefore $$2 \times 2$$ or $$4$$.

This tells us, by linearity that the determinant of the sum of these two matrices,

$\begin{pmatrix} 3 & 4 \\ 5 & 7 \end{pmatrix}$

is $$-3 + 4$$ or $$1$$. We can verify this by subtracting $$\frac{5}{3}$$ of the first row from the second.row, turning that second row into $$(0, \frac{1}{3})$$, and the product of the diagonal elements is $$1$$. You can apply linearity of the determinant in this way when two arrays have the same base, and differ only in the non-base row.

This procedure for evaluating determinants (which is sometimes called "row reduction" and sometimes called "Gaussian elimination") used on the two matrices can be applied to square arrays of any size. It is easy to do for $$2$$ by $$2$$ arrays, but it is quite easy to make a mistake even for such. It is still reasonably easy for $$3$$ by $$3$$'s but most people will make some silly mistake along the way since the steps involved in doing it are so boring, and they will get it wrong most of the time. Even you and I can expect to get $$4$$ by $$4$$ determinants wrong most of the time when doing it by hand by this approach, again because the steps are so straightforward and uninteresting. Your mind will stray along the way and you stand an excellent chance of screwing up.

Is this the only way to evaluate a determinant?

No, there are at least two other ways, one of which is equally boring and prone to your making errors. The other is magical and great fun, but surprisingly it is never taught, and few have ever heard of it.

One standard approach is to write a formula for the results of the method just described. If you start with rows $$(a, b)$$ and $$(c, d)$$. To turn the $$c$$ into $$0$$ you subtract $$\frac{c}{a}$$ times the first row from the second. The resulting diagonal elements are then $$a$$ and $$d-\frac{c}{a}$$ and their product is $$ad - bc$$. This is the formula for the determinant of a general two by two array. A standard way to calculate three by three determinants is to take the product of entries on the three downward sloping diagonals, and subtracting from their sum the sum of the products of entries on each of the three upward sloping diagonals.

Exercise 15.6 Evaluate the following determinants by any method above.

$\begin{pmatrix} 5 \end{pmatrix}, \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 8 \\ 1 & 4 & 1 \end{pmatrix}$

So what is the magical approach?