Chapter 1

Section 1.2Echelon Form and Gauss-Jordan

Elimination

Echelon Form of a MatrixA matrix that satisfies the following three conditions is said to be in echelon form.1. Every row of all zeros is at the bottom.2. The first nonzero entry from the left in

each row is a 1 (called the leading 1).3. All entries below the leading 1 in that

column are zero.

[100∗00

∗10

∗∗0

∗∗0 ] [0000

0

10000

∗1000]

[100∗00

∗00

∗00

∗10

∗∗0

∗∗1

∗∗∗

∗∗∗]

Reduced Echelon Form of a MatrixA matrix is in reduced echelon form if it is in echelon form (i.e. all entries below the leading 1 in each column are zero) and all entries above the leading 1 in each column are zero.

[100∗00

010

∗∗0

∗∗0 ] [0000

0

10000

01000]

[100∗00

∗00

∗00

010

∗∗0

001

∗∗∗

∗∗∗]

Changing Matrix FormsTo change the form a matrix is in use the row operations from the Gauss-Jordan method by starting with the top row and working your way down row by row.1. If the leading entry is zero swap with a lower row to get a nonzero number.2. Divide the row by the nonzero number to get the leading 1.3. Clear out the column below the leading 1 (above for reduced echelon) by multiplying the

current row by the negative value in the position to be cleared and adding it to that row.

{ 3 𝑥1−6𝑥2+9 𝑥3−3 𝑥5=18−𝑥1+2𝑥2−3𝑥3+2𝑥4+11𝑥5=22𝑥1−4 𝑥2+6 𝑥3−2𝑥4−12𝑥5=4

ExampleFind the augmented matrix for the system of equations to the right and row reduce it to reduced echelon form.

[ 3−12 −62−4

9−36

02−2

−311−12|

1824 ]

[ 1−12 −22−4

3−36

02−2

−111−12|

624 ]

⅓R1

[ 100 −200

300

02−2

−110−10|

68−8]

R1+R2

-2R1+R2

½R2

[ 100 −200

300

01−2

−15−10|

64−8]

2R2+R3

[ 100 −200

300

010

−150 |640]

Rewriting the equivalent system:

We can rewrite this so that the variables whose coefficient is the leading 1 in row of the matrix in reduced echelon form is on the right side of the equation and all other variables are on the left.

{𝑥1=6+2𝑥2−3𝑥3+𝑥5𝑥4=4−5 𝑥5

Find the solution for:

𝑥1=6𝑥2=1𝑥3=0𝑥4=14𝑥5=−2

General Solution

Particular Solution

Consistent and Inconsistent SystemsIf a system of equations has at least one solution (maybe more) we say the system is consistent. If no solution for a system of equations exists we say it is inconsistent.

The echelon (or reduced) form of an augmented matrix for an inconsistent system will have a leading 1 in the last (or augmented) column of the matrix.

[ 1 −4 30 0 10 0 0|

020 ] {𝑥1=4 𝑥2−3 𝑥3𝑥3=−2

ConsistentSystem

[0 1 −20 0 00 0 0

0 31 20 1] {𝑥2=3+2𝑥3𝑥4=2

𝟎=𝟏

InconsistentSystem

Having a leading 1 in the last column corresponds to the equation 0=1 which does not have any solution (i.e. it is always false) no matter what values you assign to the variables.

Row Equivalent MatricesSince the 4 basic row operations do not change the solution to the corresponding system of equations if you can transform one matrix into another by a series of row operations we call the two matrices row equivalent.

ExampleFind all values for a for which the system of equations to the right will not be consistent. { 2𝑥1+𝑎𝑥2=6

(2−𝑎 )𝑥1−4 𝑥2=0

[ 2 𝑎2−𝑎 −4|60] Augmented matrix

for system

½R1

[ 1𝑎2

2−𝑎 −4|30]-(2-a)R1+R2

[1 𝑎2

0− (2−𝑎 )𝑎

2−4| 3

−3 (2−𝑎 )]

In order for this system to not be consistent It must be that the entry in the second row and second column is zero and second row third column is not zero.

− (2−𝑎 )𝑎2

−4=0

− (2−𝑎)𝑎−8=0𝑎2−2𝑎−8=0

(𝑎−4 ) (𝑎+2 )=0

To make the second row and second column zero a has to be either 4 or -2. If a is -4 then the last row last column is -18 and the system is inconsistent. If a is -2 then the last row and last column -12 and the system is inconsistent. The values are 4 and -2.