Time to come back to this!

So last time, we came up with a shorthand way to write certain equations. Namely: the equations have to have sums of multiples of variables on the left-hand side, and constants on the right-hand side. If that's the case, we can write those coefficients in an array, put a vertical line after those coefficients, and then write the constants.

We call this an

*augmented matrix* (for reasons we will see soon). But I promised I would talk more about back-substituting.

So last time (ETA: Erk, make that the time before last, guess we get to have some repeating!), we started with the augmented matrix

| 2 1 | 5 |

| 1 2 | 4 |

and ended up with the augmented matrix

| 2 1 | 5 |

| 0 3 | 3 |

We then deduced that b = 1, a = 2, through back-substitution. Specifically, we deduced that b = 1 because 3b = 3. But we could also have determined that by multiplying the second row of that matrix by 1/3:

| 2 1 | 5 |

| 0 1 | 1 |

We then could have determined a = 2 by using another elementary row operation: subtracting the second row from the first row:

| 2 0 | 4 |

| 0 1 | 1 |

and multiplying the first row by 1/2:

| 1 0 | 2 |

| 0 1 | 1 |

This now represents the equations 1a + 0b = 2, 0a + 1b = 1, or more simply, a = 2, b = 1. So even back-substitution can be done using linear algebra - specifically, using the elementary row operations.

This suggests that we now have a general idea of what we want to do with augmented matrices: try to change the stuff before the vertical line until it has 1s on the diagonal, and 0s everywhere else. And we now have a standard process to do it (called Gauss-Jordan elimination), in 2 major steps: elimination and back-substitution:

Gaussian elimination:

1) If the top left element is zero, switch a row that has left-most element nonzero to the top.

2) Subtract multiples of the top row so the left column is otherwise zero

3) Repeat, disregarding both the first row and column (i.e. looking at terms in the second row and beyond, in the second column and beyond).

Back-substitution:

1) Multiply the last row so that it has a 1 first.

2) Subtract multiples of the last row from rows above it so the right column is otherwise zero.

3) Repeat, disregarding both the last row and column.

Next time, we talk about a somewhat more general possibility - where we can't always get that 1s and 0s matrix.