math 1300: section 4-2 systems of linear equations; augmented matrices

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MatricesSolving Linear Systems Using Augmented Matrices

SummaryReminders

Math 1300 Finite MathematicsSection 4.2 Systems of Linear Equations and Augmented

Matrices

Department of MathematicsUniversity of Missouri

September 16, 2009

Math 1300 Finite Mathematics

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SummaryReminders

A matrix is a rectangular array of numbers written with brackets.

1 −9 3−1 .5 63 −4 5

Each number in a matrix is called an element of the matrix.If a matrix has m rows and n columns, it is called an m × nmatrix.

The matrix above is 3× 3. A 3× 1 matrix:

−13−6


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SummaryReminders

A matrix is a rectangular array of numbers written with brackets. 1 −9 3−1 .5 63 −4 5



−13−6


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SummaryReminders


Each number in a matrix is called an element of the matrix.

If a matrix has m rows and n columns, it is called an m × nmatrix.


−13−6


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SummaryReminders




−13−6


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SummaryReminders

The expression m × n is called the size of the matrix

The numbers m and n are called the dimensions of thematrix.A matrix with n rows and n columns is called a squarematrix of order n.A square matrix of order 3: 1 −9 3

−1 .5 63 −4 5


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SummaryReminders

The expression m × n is called the size of the matrixThe numbers m and n are called the dimensions of thematrix.

A matrix with n rows and n columns is called a squarematrix of order n.A square matrix of order 3: 1 −9 3

−1 .5 63 −4 5


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SummaryReminders

The expression m × n is called the size of the matrixThe numbers m and n are called the dimensions of thematrix.A matrix with n rows and n columns is called a squarematrix of order n.

A square matrix of order 3: 1 −9 3−1 .5 63 −4 5


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SummaryReminders

The expression m × n is called the size of the matrixThe numbers m and n are called the dimensions of thematrix.A matrix with n rows and n columns is called a squarematrix of order n.A square matrix of order 3: 1 −9 3

−1 .5 63 −4 5


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SummaryReminders

A matrix with only one column is called a column matrix.

−13−6

is a column matrix.

A matrix with only one row is called a row matrix.[3 −4 2

]is a row matrix.


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SummaryReminders

A matrix with only one column is called a column matrix.−13−6

is a column matrix.


]is a row matrix.


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SummaryReminders


is a column matrix.

A matrix with only one row is called a row matrix.

[3 −4 2

]is a row matrix.


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SummaryReminders


is a column matrix.


]is a row matrix.


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SummaryReminders

The position of an element in a matrix is given by the rowand column containing the element.

This is usually denoted using double subscript notation aijwhere i is the row and j is the column containing theelement aij .

For example, given

1 −9 3−1 .5 63 −4 5

then a11 = 1, a21 = −1,

a23 = 6, a32 = −4, etc.The principal diagonal of a matrix is made up of theelements a11, a22, a33, etc.


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SummaryReminders

The position of an element in a matrix is given by the rowand column containing the element.This is usually denoted using double subscript notation aijwhere i is the row and j is the column containing theelement aij .

For example, given

1 −9 3−1 .5 63 −4 5

then a11 = 1, a21 = −1,



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SummaryReminders


For example, given

1 −9 3−1 .5 63 −4 5

then a11 = 1,

a21 = −1,



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SummaryReminders


For example, given

1 −9 3−1 .5 63 −4 5

then a11 = 1, a21 = −1,



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SummaryReminders


For example, given

1 −9 3−1 .5 63 −4 5

then a11 = 1, a21 = −1,

a23 = 6,

a32 = −4, etc.The principal diagonal of a matrix is made up of theelements a11, a22, a33, etc.


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SummaryReminders


For example, given

1 −9 3−1 .5 63 −4 5

then a11 = 1, a21 = −1,

a23 = 6, a32 = −4, etc.

The principal diagonal of a matrix is made up of theelements a11, a22, a33, etc.


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SummaryReminders


For example, given

1 −9 3−1 .5 63 −4 5

then a11 = 1, a21 = −1,



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SummaryReminders

Related to a linear system are several important matrices:

thecoefficient matrix, the constant matrix, and the augmentedmatrix. For example, given

x1 − 4x2 = −2−2x1 + x2 = 10

we have:

The coefficient matrix:[

1 −4−2 1

]The constant matrix:

[−210

]The augmented matrix:

[1 −4−2 1

∣∣∣∣−210

]


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SummaryReminders

Related to a linear system are several important matrices: thecoefficient matrix,

the constant matrix, and the augmentedmatrix. For example, given

x1 − 4x2 = −2−2x1 + x2 = 10

we have:


1 −4−2 1


[−210


[1 −4−2 1

∣∣∣∣−210

]


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SummaryReminders

Related to a linear system are several important matrices: thecoefficient matrix, the constant matrix,

and the augmentedmatrix. For example, given

x1 − 4x2 = −2−2x1 + x2 = 10

we have:


1 −4−2 1


[−210


[1 −4−2 1

∣∣∣∣−210

]


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SummaryReminders

Related to a linear system are several important matrices: thecoefficient matrix, the constant matrix, and the augmentedmatrix.

For example, given

x1 − 4x2 = −2−2x1 + x2 = 10

we have:


1 −4−2 1


[−210


[1 −4−2 1

∣∣∣∣−210

]


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SummaryReminders

Related to a linear system are several important matrices: thecoefficient matrix, the constant matrix, and the augmentedmatrix. For example, given

x1 − 4x2 = −2−2x1 + x2 = 10

we have:


1 −4−2 1


[−210


[1 −4−2 1

∣∣∣∣−210

]


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10

we have:


1 −4−2 1

]

The constant matrix:[−210


[1 −4−2 1

∣∣∣∣−210

]


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10

we have:


1 −4−2 1


[−210

]

The augmented matrix:[

1 −4−2 1

∣∣∣∣−210

]


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10

we have:


1 −4−2 1


[−210


[1 −4−2 1

∣∣∣∣−210

]


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SummaryReminders

Recall that two linear systems are equivalent if they havethe same solution set.

Similarly, two augmented matrices are row equivalent ifthey are augmented matrices of equivalent systems oflinear equations.


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SummaryReminders

Recall that two linear systems are equivalent if they havethe same solution set.Similarly, two augmented matrices are row equivalent ifthey are augmented matrices of equivalent systems oflinear equations.


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SummaryReminders

Theorem (Operations that Produce Row-Equivalent Matrices)An augmented matrix is transformed into a row-equivalentmatrix by performing any of the following row operations:

(A) Two rows are interchanged(B) A row is multiplied by a nonzero constant(C) A constant multiple of one row is added to another row.


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SummaryReminders

Theorem (Operations that Produce Row-Equivalent Matrices)An augmented matrix is transformed into a row-equivalentmatrix by performing any of the following row operations:(A) Two rows are interchanged

(B) A row is multiplied by a nonzero constant(C) A constant multiple of one row is added to another row.


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SummaryReminders

Theorem (Operations that Produce Row-Equivalent Matrices)An augmented matrix is transformed into a row-equivalentmatrix by performing any of the following row operations:(A) Two rows are interchanged(B) A row is multiplied by a nonzero constant

(C) A constant multiple of one row is added to another row.


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SummaryReminders

Theorem (Operations that Produce Row-Equivalent Matrices)An augmented matrix is transformed into a row-equivalentmatrix by performing any of the following row operations:(A) Two rows are interchanged(B) A row is multiplied by a nonzero constant(C) A constant multiple of one row is added to another row.


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SummaryReminders

Example: Which row operation was used to transform[−1 26 −3

∣∣∣∣−312

]to [

−1 22 −1

∣∣∣∣−34

]?

13R2


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SummaryReminders


∣∣∣∣−312

]to [

6 −3−1 2

∣∣∣∣12−3

]

R1 ↔ R2


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SummaryReminders


∣∣∣∣−312

]to [

6 −3−1 2

∣∣∣∣12−3

]R1 ↔ R2


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SummaryReminders


∣∣∣∣−312

]to [

−1 22 5

∣∣∣∣−30

]

4R1 + R2 → R2


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SummaryReminders

Example: Solve the system of linear equations

x1 − 4x2 = −2−2x1 + x2 = 10

using augmented matrix methods.

[1 −4−2 1

∣∣∣∣−210

]2R1+R2→R2−−−−−−−−→

[1 −40 −7

∣∣∣∣−26

]− 1

7 R2→R2−−−−−−→[1 −40 1

∣∣∣∣−2−6

7

]4R2+R1→R1−−−−−−−−→

[1 00 1

∣∣∣∣−387−6

7

]Therefore the solution is x1 = −38

7 , x2 = −67 .


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10

using augmented matrix methods.[1 −4−2 1

∣∣∣∣−210

]

2R1+R2→R2−−−−−−−−→[1 −40 −7

∣∣∣∣−26

]− 1

7 R2→R2−−−−−−→[1 −40 1

∣∣∣∣−2−6

7

]4R2+R1→R1−−−−−−−−→

[1 00 1

∣∣∣∣−387−6

7


7 , x2 = −67 .


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10


∣∣∣∣−210

]2R1+R2→R2−−−−−−−−→

[1 −40 −7

∣∣∣∣−26

]− 1

7 R2→R2−−−−−−→[1 −40 1

∣∣∣∣−2−6

7

]4R2+R1→R1−−−−−−−−→

[1 00 1

∣∣∣∣−387−6

7


7 , x2 = −67 .


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10


∣∣∣∣−210

]2R1+R2→R2−−−−−−−−→

[1 −40 −7

∣∣∣∣−26

]

− 17 R2→R2−−−−−−→

[1 −40 1

∣∣∣∣−2−6

7

]4R2+R1→R1−−−−−−−−→

[1 00 1

∣∣∣∣−387−6

7


7 , x2 = −67 .


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10


∣∣∣∣−210

]2R1+R2→R2−−−−−−−−→

[1 −40 −7

∣∣∣∣−26

]− 1

7 R2→R2−−−−−−→

[1 −40 1

∣∣∣∣−2−6

7

]4R2+R1→R1−−−−−−−−→

[1 00 1

∣∣∣∣−387−6

7


7 , x2 = −67 .


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10


∣∣∣∣−210

]2R1+R2→R2−−−−−−−−→

[1 −40 −7

∣∣∣∣−26

]− 1

7 R2→R2−−−−−−→[1 −40 1

∣∣∣∣−2−6

7

]

4R2+R1→R1−−−−−−−−→[1 00 1

∣∣∣∣−387−6

7


7 , x2 = −67 .


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10


∣∣∣∣−210

]2R1+R2→R2−−−−−−−−→

[1 −40 −7

∣∣∣∣−26

]− 1

7 R2→R2−−−−−−→[1 −40 1

∣∣∣∣−2−6

7

]4R2+R1→R1−−−−−−−−→

[1 00 1

∣∣∣∣−387−6

7


7 , x2 = −67 .


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10


∣∣∣∣−210

]2R1+R2→R2−−−−−−−−→

[1 −40 −7

∣∣∣∣−26

]− 1

7 R2→R2−−−−−−→[1 −40 1

∣∣∣∣−2−6

7

]4R2+R1→R1−−−−−−−−→

[1 00 1

∣∣∣∣−387−6

7

]

Therefore the solution is x1 = −387 , x2 = −6

7 .


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SummaryReminders


x1 − 4x2 = −2−2x1 + x2 = 10


∣∣∣∣−210

]2R1+R2→R2−−−−−−−−→

[1 −40 −7

∣∣∣∣−26

]− 1

7 R2→R2−−−−−−→[1 −40 1

∣∣∣∣−2−6

7

]4R2+R1→R1−−−−−−−−→

[1 00 1

∣∣∣∣−387−6

7


7 , x2 = −67 .


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SummaryReminders


0.2x1 − 0.5x2 = 0.070.8x1 − 0.3x2 = 0.79

using augmented matrix methods.

[0.2 −0.50.8 −0.3

∣∣∣∣0.070.79

] 10.2 R1→R1−−−−−−→

[1 −2.5

0.8 −0.3

∣∣∣∣0.350.79

]−0.8R1+R2→R2−−−−−−−−−−→[

1 −2.50 1.7

∣∣∣∣0.350.51

]1

1.7 R2−−−→[1 −2.50 1

∣∣∣∣0.350.3

]2.5R2+R1→R1−−−−−−−−−→[

1 00 1

∣∣∣∣1.10.3

]So, x1 = 1.1, x2 = 0.3


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SummaryReminders


0.2x1 − 0.5x2 = 0.070.8x1 − 0.3x2 = 0.79

using augmented matrix methods.[0.2 −0.50.8 −0.3

∣∣∣∣0.070.79

]

10.2 R1→R1−−−−−−→

[1 −2.5

0.8 −0.3

∣∣∣∣0.350.79

]−0.8R1+R2→R2−−−−−−−−−−→[

1 −2.50 1.7

∣∣∣∣0.350.51

]1

1.7 R2−−−→[1 −2.50 1

∣∣∣∣0.350.3

]2.5R2+R1→R1−−−−−−−−−→[

1 00 1

∣∣∣∣1.10.3

]So, x1 = 1.1, x2 = 0.3


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SummaryReminders


0.2x1 − 0.5x2 = 0.070.8x1 − 0.3x2 = 0.79


∣∣∣∣0.070.79

] 10.2 R1→R1−−−−−−→

[1 −2.5

0.8 −0.3

∣∣∣∣0.350.79

]−0.8R1+R2→R2−−−−−−−−−−→[

1 −2.50 1.7

∣∣∣∣0.350.51

]1

1.7 R2−−−→[1 −2.50 1

∣∣∣∣0.350.3

]2.5R2+R1→R1−−−−−−−−−→[

1 00 1

∣∣∣∣1.10.3

]So, x1 = 1.1, x2 = 0.3


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SummaryReminders


0.2x1 − 0.5x2 = 0.070.8x1 − 0.3x2 = 0.79


∣∣∣∣0.070.79

] 10.2 R1→R1−−−−−−→

[1 −2.5

0.8 −0.3

∣∣∣∣0.350.79

]

−0.8R1+R2→R2−−−−−−−−−−→[1 −2.50 1.7

∣∣∣∣0.350.51

]1

1.7 R2−−−→[1 −2.50 1

∣∣∣∣0.350.3

]2.5R2+R1→R1−−−−−−−−−→[

1 00 1

∣∣∣∣1.10.3

]So, x1 = 1.1, x2 = 0.3


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SummaryReminders


0.2x1 − 0.5x2 = 0.070.8x1 − 0.3x2 = 0.79


∣∣∣∣0.070.79

] 10.2 R1→R1−−−−−−→

[1 −2.5

0.8 −0.3

∣∣∣∣0.350.79

]−0.8R1+R2→R2−−−−−−−−−−→

[1 −2.50 1.7

∣∣∣∣0.350.51

]1

1.7 R2−−−→[1 −2.50 1

∣∣∣∣0.350.3

]2.5R2+R1→R1−−−−−−−−−→[

1 00 1

∣∣∣∣1.10.3

]So, x1 = 1.1, x2 = 0.3


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SummaryReminders


0.2x1 − 0.5x2 = 0.070.8x1 − 0.3x2 = 0.79


∣∣∣∣0.070.79

] 10.2 R1→R1−−−−−−→

[1 −2.5

0.8 −0.3

∣∣∣∣0.350.79

]−0.8R1+R2→R2−−−−−−−−−−→[

1 −2.50 1.7

∣∣∣∣0.350.51

]1

1.7 R2−−−→

[1 −2.50 1

∣∣∣∣0.350.3

]2.5R2+R1→R1−−−−−−−−−→[

1 00 1

∣∣∣∣1.10.3

]So, x1 = 1.1, x2 = 0.3


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SummaryReminders


0.2x1 − 0.5x2 = 0.070.8x1 − 0.3x2 = 0.79


∣∣∣∣0.070.79

] 10.2 R1→R1−−−−−−→

[1 −2.5

0.8 −0.3

∣∣∣∣0.350.79

]−0.8R1+R2→R2−−−−−−−−−−→[

1 −2.50 1.7

∣∣∣∣0.350.51

]1

1.7 R2−−−→[1 −2.50 1

∣∣∣∣0.350.3

]

2.5R2+R1→R1−−−−−−−−−→[1 00 1

∣∣∣∣1.10.3

]So, x1 = 1.1, x2 = 0.3


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SummaryReminders


0.2x1 − 0.5x2 = 0.070.8x1 − 0.3x2 = 0.79


∣∣∣∣0.070.79

] 10.2 R1→R1−−−−−−→

[1 −2.5

0.8 −0.3

∣∣∣∣0.350.79

]−0.8R1+R2→R2−−−−−−−−−−→[

1 −2.50 1.7

∣∣∣∣0.350.51

]1

1.7 R2−−−→[1 −2.50 1

∣∣∣∣0.350.3

]2.5R2+R1→R1−−−−−−−−−→

[1 00 1

∣∣∣∣1.10.3

]So, x1 = 1.1, x2 = 0.3


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SummaryReminders


0.2x1 − 0.5x2 = 0.070.8x1 − 0.3x2 = 0.79


∣∣∣∣0.070.79

] 10.2 R1→R1−−−−−−→

[1 −2.5

0.8 −0.3

∣∣∣∣0.350.79

]−0.8R1+R2→R2−−−−−−−−−−→[

1 −2.50 1.7

∣∣∣∣0.350.51

]1

1.7 R2−−−→[1 −2.50 1

∣∣∣∣0.350.3

]2.5R2+R1→R1−−−−−−−−−→[

1 00 1

∣∣∣∣1.10.3

]

So, x1 = 1.1, x2 = 0.3


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SummaryReminders


0.2x1 − 0.5x2 = 0.070.8x1 − 0.3x2 = 0.79


∣∣∣∣0.070.79

] 10.2 R1→R1−−−−−−→

[1 −2.5

0.8 −0.3

∣∣∣∣0.350.79

]−0.8R1+R2→R2−−−−−−−−−−→[

1 −2.50 1.7

∣∣∣∣0.350.51

]1

1.7 R2−−−→[1 −2.50 1

∣∣∣∣0.350.3

]2.5R2+R1→R1−−−−−−−−−→[

1 00 1

∣∣∣∣1.10.3

]So, x1 = 1.1, x2 = 0.3


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SummaryReminders

Example Solve the system using augmented matrix methods:

2x1 − x2 = 4−6x1 + 3x2 = −12

[2 −1−6 3

∣∣∣∣ 4−12

]12 R1→R1−−−−−→

[1 −1

2−6 3

∣∣∣∣ 2−12

]13 R2→R2−−−−−→

[1 −1

2−2 1

∣∣∣∣ 2−4

]2R1+R2→R2−−−−−−−−→

[1 −1

20 0

∣∣∣∣20]


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SummaryReminders


2x1 − x2 = 4−6x1 + 3x2 = −12

[2 −1−6 3

∣∣∣∣ 4−12

]

12 R1→R1−−−−−→

[1 −1

2−6 3

∣∣∣∣ 2−12

]13 R2→R2−−−−−→

[1 −1

2−2 1

∣∣∣∣ 2−4

]2R1+R2→R2−−−−−−−−→

[1 −1

20 0

∣∣∣∣20]


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SummaryReminders


2x1 − x2 = 4−6x1 + 3x2 = −12

[2 −1−6 3

∣∣∣∣ 4−12

]12 R1→R1−−−−−→

[1 −1

2−6 3

∣∣∣∣ 2−12

]13 R2→R2−−−−−→

[1 −1

2−2 1

∣∣∣∣ 2−4

]2R1+R2→R2−−−−−−−−→

[1 −1

20 0

∣∣∣∣20]


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SummaryReminders


2x1 − x2 = 4−6x1 + 3x2 = −12

[2 −1−6 3

∣∣∣∣ 4−12

]12 R1→R1−−−−−→

[1 −1

2−6 3

∣∣∣∣ 2−12

]

13 R2→R2−−−−−→

[1 −1

2−2 1

∣∣∣∣ 2−4

]2R1+R2→R2−−−−−−−−→

[1 −1

20 0

∣∣∣∣20]


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SummaryReminders


2x1 − x2 = 4−6x1 + 3x2 = −12

[2 −1−6 3

∣∣∣∣ 4−12

]12 R1→R1−−−−−→

[1 −1

2−6 3

∣∣∣∣ 2−12

]13 R2→R2−−−−−→

[1 −1

2−2 1

∣∣∣∣ 2−4

]2R1+R2→R2−−−−−−−−→

[1 −1

20 0

∣∣∣∣20]


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SummaryReminders


2x1 − x2 = 4−6x1 + 3x2 = −12

[2 −1−6 3

∣∣∣∣ 4−12

]12 R1→R1−−−−−→

[1 −1

2−6 3

∣∣∣∣ 2−12

]13 R2→R2−−−−−→

[1 −1

2−2 1

∣∣∣∣ 2−4

]

2R1+R2→R2−−−−−−−−→[1 −1

20 0

∣∣∣∣20]


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SummaryReminders


2x1 − x2 = 4−6x1 + 3x2 = −12

[2 −1−6 3

∣∣∣∣ 4−12

]12 R1→R1−−−−−→

[1 −1

2−6 3

∣∣∣∣ 2−12

]13 R2→R2−−−−−→

[1 −1

2−2 1

∣∣∣∣ 2−4

]2R1+R2→R2−−−−−−−−→

[1 −1

20 0

∣∣∣∣20]


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SummaryReminders

The last matrix [1 −1

20 0

∣∣∣∣20]

corresponds to the system

x1 −12

x2 = 2

0 = 0

This system is equivalent to the original system.Geometrically, the graphs of the two original equationscoincide and there are infinitely many solutions.We represent the infinitely many solutions by introducing aparameter.


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SummaryReminders


20 0

∣∣∣∣20]


x1 −12

x2 = 2

0 = 0

This system is equivalent to the original system.Geometrically, the graphs of the two original equationscoincide and there are infinitely many solutions.

We represent the infinitely many solutions by introducing aparameter.


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SummaryReminders


20 0

∣∣∣∣20]


x1 −12

x2 = 2

0 = 0

This system is equivalent to the original system.Geometrically, the graphs of the two original equationscoincide and there are infinitely many solutions.We represent the infinitely many solutions by introducing aparameter.


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SummaryReminders

To do so, we first solve

x1 =12

x2 + 2

We now introduce our parameter t , and set x2 = t .The solution may now be stated: The solution set is the setof all pairs (x1, x2) where

x1 =12

t + 2

x2 = t

for some real number t .


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SummaryReminders


x1 =12

x2 + 2

We now introduce our parameter t , and set x2 = t .

The solution may now be stated: The solution set is the setof all pairs (x1, x2) where

x1 =12

t + 2

x2 = t



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SummaryReminders


x1 =12

x2 + 2

We now introduce our parameter t , and set x2 = t .The solution may now be stated: The solution set is the setof all pairs (x1, x2) where

x1 =12

t + 2

x2 = t



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SummaryReminders

Example: Solve the system using augmented matrix methods:

2x1 + 6x2 = −3x1 + 3x2 = 2

[2 61 3

∣∣∣∣−32

]− 1

2 R1+R2→R2−−−−−−−−−→[2 60 0

∣∣∣∣−352

]This is the augmented matrix of the system

2x1 + 6x2 = −3

0 =52

Therefore the original system is inconsistent, and has nosolution.


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SummaryReminders


2x1 + 6x2 = −3x1 + 3x2 = 2[

2 61 3

∣∣∣∣−32

]

− 12 R1+R2→R2−−−−−−−−−→

[2 60 0

∣∣∣∣−352


2x1 + 6x2 = −3

0 =52



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SummaryReminders


2x1 + 6x2 = −3x1 + 3x2 = 2[

2 61 3

∣∣∣∣−32

]− 1

2 R1+R2→R2−−−−−−−−−→

[2 60 0

∣∣∣∣−352


2x1 + 6x2 = −3

0 =52



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SummaryReminders


2x1 + 6x2 = −3x1 + 3x2 = 2[

2 61 3

∣∣∣∣−32

]− 1

2 R1+R2→R2−−−−−−−−−→[2 60 0

∣∣∣∣−352

]

This is the augmented matrix of the system

2x1 + 6x2 = −3

0 =52



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SummaryReminders


2x1 + 6x2 = −3x1 + 3x2 = 2[

2 61 3

∣∣∣∣−32

]− 1

2 R1+R2→R2−−−−−−−−−→[2 60 0

∣∣∣∣−352


2x1 + 6x2 = −3

0 =52



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SummaryReminders

Summary

Form 1: A Unique Solution (Consistent and Independent)[1 00 1

∣∣∣∣mn]

Form 2: Infinitely Many Solutons (Consistent and Dependent)[1 m0 0

∣∣∣∣n0]

Form 3: No Solution (Inconsistent)[1 m0 0

∣∣∣∣np]

p 6= 0


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SummaryReminders

Reminders

Written HW: 4.1: 11, 14, 26, 54 and 4.2: 13, 27, 45, 53, 65Online HW: Online HW 3 Due 9/17 8amExam 1: Wednesday Sept 23 6:30-7:30pm - Section 24 isin Waters Auditorium.


math 1300: section 4-2 systems of linear equations; augmented matrices

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