On the previous page we learned how to evaluate a 2 x 2 determinant, but of course determinants can be larger than 2 x 2, so let's examine how to evaluate a 3 x 3 determinant.

= aei + dhc + gbf - gec - dbi - ahf.

Now, that probably looks horrific and an awful mess to remember, but there's a simple way of remembering what we multiply to get the value.

Notice that aei is the diagonal starting from the top left, going to the bottom right.

dhc is the next diagonal down from there (when we run out of rows drawing that diagonal, we continue at the top).

Similiarly, gbf is the last diagonal going in that direction.

Now look at the ones we're subtracting: gec is the diagonal starting from the bottom left going to the top right.

dbi is the next diagonal in that direction, and ahf is the third one.

So we add the diagonal products going from top left to bottom right, and subtract the ones going from bottom left to top right.

Problem One
Calculate the value of .

Solution
1(7)(0) + 5(4)(2) + 2(3)(1) - 2(7)(2) - 5(3)(0) - 1(4)(1)
= 0 + 40 + 6 - 28 - 0 - 4
= 14

Problem Two
Calculate the value of

Solution
Be very careful here; there are some negative numbers, and we have to properly handle the sign when multiplying.

1(2)(2) + (-1)(-3)(3) + 1(2)(1) - 1(2)(3) - (-1)(2)(2) - 1(-3)(1)
= 4 + 9 + 2 - 6 +4 +3
= 16

Problem Three
Solve for x if = 13

Solution
x(2)(3) + 2(1)(1) + 1(1)(0) - 1(3)(1) - 2(1)(2) - x(1)(0) = 13
6x + 2 - 3 - 4 = 13
6x = 18
x = 3

1.

Find the value of

2.

Find the value of

3.

Find the value of

4.

Find x if = 10

5.

Find x if = 2