Math 2940 - Week 1 Recitation
Review
Given a linear system of equations, the
augmented matrix
of the system is the matrix
containing the coefficients and constant values of each equation in rows, with each variable
corresponding to a column.
There are three elementary row operations that can be applied to an augmented matrix.
These operations do not change the solution set of the associated linear system.
1.
Replacement: replace a row with the sum of itself and a multiple of another row.
2.
Interchange: swap two rows.
3.
Scaling: multiply all entries in a row by a nonzero constant.
A matrix is in
echelon
form if
1.
All rows of all zeros are at the bottom.
2.
Any leading entry is strictly to the right of the leading entry in the row above it.
3.
All entries in a column below a leading entry are zero.
and is in
reduced echelon
form if in addition
4.
The leading entry in each nonzero row is 1.
5.
Each leading 1 is the only nonzero entry in its column.
Linear systems in reduced echelon form are easy to solve, and the row reduction algorithm
(sometimes called Gaussian elimination) puts any matrix in reduced echelon form.
1
Math 2940 - Week 1 Recitation
Problems
1. Find the solution set to the linear system
3x − 4y + z = 1
−9x + 10y − 6z = 3
−6x + 10y − 4z = 0
using row reduction of the augmented matrix.
The augmented matrix is


3 −4 1 1
−9 10 −6 3 .
−6 10 −4 0
In performing row reduction, we first pick the top left corner as pivot and zero out the
leftmost column by adding three times the first row to the second row and two times
the first row to the third row to get


3 −4 1 1
0 −2 −3 6 .
0 2 −2 2
Then we pick the second entry in the second column as the pivot and zero out the
column below by adding one times the second row to the third. The result is


3 −4 1 1
0 −2 −3 6 .
0 0 −5 8
The matrix is now in echelon form. We can put it in reduced echelon form by dividing
each row by its leading coefficient to get


1 − 34 31 13
0 1 3 −3 
2
0 0 1 − 85
and then clearing each column (subtract 32 of the third row from the second and subtract
−4
of the second and 31 of the third from the first row) with the pivots to get
3


1
1 0 0 15
0 1 0 − 3  .
5
0 0 1 − 85
The associated linear system is
1
15
3
y=−
5
8
z=−
5
x=
2
Math 2940 - Week 1 Recitation
which directly gives the complete solution set
(x, y, z) =
3 8
1
,− ,−
15 5 5
.
2. Find the solution set to the linear system
2x1 + 6x2 − 2x3 = 14
2x1 + 6x2 + x4 = 3
3x1 + 9x2 − x3 + x4 = 10
using row reduction of the augmented matrix.
The augmented matrix is


2 6 −2 0 14
2 6 0 1 3 
3 9 −1 1 10
which reduces to (make sure you can fill in the steps of the row reduction I’m skipping
here)


3
1 3 0 21
2
0 0 1 1 − 11  .
2
2
0 0 0 0 0
This corresponds to the system
1
3
x1 + 3x2 + x4 =
2
2
11
1
x3 + = −
2
2
that we note has free variables x2 and x4 . Thus the solution set is
3
1
− 3x2 − x4
2
2
x2 is free
11 1
x3 = − − x 4
2
2
x4 is free.
x1 =
3