One more thing about 3x3 determinants

While we're on the topic of determinants (see yesterday's post), rather than pounding on arithmetic drills, the other thing I would show algebra students is that you can use any row or column to set up the computation. That can be useful if one choice contains zeros.

The standard approach uses the top row:

a b c d e f g h i

And the determinant is:

a(ei-hf) - b(di-fg) + c(dh-eg)

Where the sign of the second term is negative by the checkerboard rule:

+ - + - + - + - +

Multiplying out we obtain:

aei - afh - bdi + bfg + cdh - ceg

The 6 terms contain three components, each taken from a different row and column. For example, the components of bdi are from:

b = (1,2) d = (2,1) i = (3,3)

The checkerboard rule makes the sign come out correctly.

If we're working with the top row or the middle column and so processing b x (di - fg) or d x (bi - ch), we'll need the minus sign; whereas if we're obtaining this term from i x (ae - bd) we already have a minus sign.

Let's try using the last row. We have:

a b c d e f g h i g(bf - ce) - h(af - cd) + i(ae - bd) bfg - cdg - afh + cdh + aei - bdi

Compare with the first example to see that all the terms are present.

Can we do it by the diagonal? Try ceg:

c(dh - eg) - e(ai - cg) + g(bf - ce)

Nope. Some terms are correct, but some are duplicates.