Page 59 - A Course in Linear Algebra with Applications

2.2: Elementary Row Operations 43

Then we put M in row echelon form, using row operations.
From this we can determine if the original linear system is
consistent; for this to be true, in the row echelon form of M
the scalars in the last column which lie below the final pivot
must all be zero. To find the general solution of a consistent
system we convert the row echelon matrix back to a linear
system and use back substitution to solve it.

Example 2.2.2
Consider once again the linear system of Example 2.1.3;

Xl + 3x 2 + 3x 3 + 2x 4 = 1
+ 5X4 = 5
2xi + 6x 2 + 9x 3
-Xi - 3x 2 + 3x 3 = 5

The augmented matrix here is

1 3 3 2 1 1
2 6 9 5 1 5
1 — 3 3 0 1 5

Now we have just seen in Example 2.2.1 that this matrix has
row echelon form

1 3 3 2 1 1
0 0 11/3 | 1
0 0 0 0 1 0

Because the lower right hand entry is 0, the linear system is
consistent. The linear system corresponding to the last matrix
is

Xi + 3X2 + 3X3 + 2X4 = 1

£3 + - x 4 = 1
0 = 0