# Matrices

# Cramer's Rule Redux

## Determinants and Cramer's Rule Redux: Because Things Happen in Threes

Here is the first trick for using Cramer's Rule on 3 × 3 matrices. This does not work for 4 × 4s or larger matrices, and of course we don't need it for 2 × 2s.

Let's say we have this saucy little 3 × 3 matrix that we'd just love to introduce to our friend Cramer, if only she would bring her friend Determinant along. Here's what we do.

First, copy the left and center columns to the right of the matrix:

You remember, of course, that the 2 × 2 Cramer's technique had us multiply diagonally from upper left to lower right, and then subtract from lower left to upper right. This is actually pretty similar. The main thing is to remember not to mix up your columns that you copy again. Now that we have our matrix and the left and center columns from it copied outside the bars, we're ready to calculate the determinate in the same way that we did the 2 × 2.

We first multiply from the top left corner to the bottom right, and then add. We do that three times—that's why we added the first two columns again:

D = (a)(e)(i) + (b)(f)(g) + (c)(d)(h)…

and then we subtract up from bottom left to upper right again, just like we did before:

D = (a)(e)(i) + (b)(f)(g) + (c)(d)(h) – (g)(e)(c) – (h)(f)(a) – (i)(d)(b)

and that's it.

So in a real numbers situation:

we first copy the first two columns just outside and to the right of the bars:

and from here on out, it's basically plug 'n chug:

D = (4)(1)(4) + (1)(5)(3) + (0)(2)(0) – (3)(1)(0) – (0)(5)(4) – (4)(2)(1)

D = 16 + 15 + 0 – 0 – 8 = 23

Got it.

Now, since we know how to get the determinant for a 3 × 3 matrix, here's how we solve for variables with Cramer's Rule when there are three of them. Here are our three lucky bachelors:

2*x + y – 3z = 8*

-3x + 2y + z = 10

x – 3y +2z = 12

The first step is to find D, the coefficient determinant. We just learned how to do that for a 3 × 3 matrix, so let's go.

First create the coefficient matrix:

Whoomp! There it is. Next, find D:

We remember our little trick, wherein we copy the first two columns again outside and to the right of the matrix:

Excellent, Smithers. Now we multiply and add diagonally across from top left to bottom right:

D = (2)(2)(2) + (1)(1)(1) + (-3)(-3)(-3)…

We then multiply and subtract diagonally across from bottom left to top right:

D = (2)(2)(2) + (1)(1)(1) + (-3)(-3)(-3) – (1)(2)(-3) – (-3)(1)(2) – (2)(-3)(1)

So:

D = 8 + 1 + (-27) – (-6) – (-6) – (-6) = 0

Ruh roh, Shaggy! When *D = 0*, there can't be any exact values! (Did you remember?) We thought those equations looked suspicious. Well, how about these?

3*x – y – 4z = 10*

-2x + 5y + z = 9

-3x + 4y + 2z = 8

First we bust out our coefficient matrix:

You down with finding D? Yeah, you know me:

We copy the first two columns right outside and to the right of the matrix:

Then it's time to multiply and add diagonally across from top left to bottom right and then multiply and subtract diagonally across from bottom left to top right:

D = (3)(5)(2) + (-1)(1)(-3) + (-4)(-2)(4) – (-3)(5)(-4) – (4)(1)(3) – (2)(-2)(-1)

From there, it's mathematics bliss:

D = 30 + 3 + 32 – 60 – 12 – 4 = -11

We know that D = -11 so now we're after those other determinants. We know that these three things are true:

We know that the way to find D_{x} is to substitute the *= values* into the mix in place of the *x values* and then use our column-copying trick to get the numbers. Time for more plug and chug. Remove the *x values*:

Replace them with the *= values*:

Copy the first two columns in our modified matrix just outside and to the right:

Multiply and add diagonally across from top left to bottom right and then multiply and subtract diagonally across from bottom left to top right:

D_{x} = (10)(5)(2) + (-1)(1)(8) + (-4)(9)(4) – (8)(5)(-4) – (4)(1)(10) – (2)(9)(-1)

D_{x} = 100 + (-8) + (-144) – (-160) – 40 – (-18) = 86

Might as well solve for *x* – we're so close:

Onward. We seek D_{y}

So

There it is, devoid of all *y* values. Now we plug in the = values and roll from there:

Column trick:

Plug and chug:

D_{y} = (3)(9)(2) + (10)(1)(-3) + (-4)(-2)(8) – (-3)(9)(-4) – (8)(1)(3) – (2)(-2)(10)

=54 + (-30) + 64 – 108 – 24 –(-40) = -4

Solve for *y*, as long as we're kickin it here:

We are *so* almost there. Here we go, our little matrix, naked of all *z values*:

Substitute in the *= values*:

Ah, *= values*. You complete us. Column trick:

Plug and chug, baby:

D_{z} = (3)(5)(8) + (-1)(9)(-3) + (10)(-2)(4) – (-3)(5)(10) – (4)(9)(3) – (8)(-2)(-1)

= 120 + 127 + (-80) – (-150) – 108 – 16 = 93

Another triumph of the human spirit. At last we solve for our *z* determinant:

So:

Like a German opera, these are time-consuming and can feel repetitive, but the payoff feels great. (And if you hate opera, or algebra, the payoff is leaving or being done, so that's even better.)