MATH1131 Mathematics 1A Algebra

UNSW Sydney Semester 1, 2017

Maike Massierer

Chapter 4: Linear Equations and Matrices

Lecture 19: Deducing solubility from row-echelon formand solving systems with indeterminate right hand side

Definition

a leading row is a nonzero row

a leading entry is the first nonzero entry in a leading row

a leading column is a column containing a leading entry

Definition

a leading row is a nonzero row

a leading entry is the first nonzero entry in a leading row

a leading column is a column containing a leading entry

a leading variable is a variable xi represented by a leading column

Definition

a leading row is a nonzero row

a leading entry is the first nonzero entry in a leading row

a leading column is a column containing a leading entry

a leading variable is a variable xi represented by a leading column

ExampleConsider the following matrix:

(

0 5 70 0 0

)

Row 1 is a leading row with leading entry 5.Row 2 is a non-leading row.Column 2 is a leading column but columns 1 and 3 are not.

Definition

a leading row is a nonzero row

a leading entry is the first nonzero entry in a leading row

a leading column is a column containing a leading entry

a leading variable is a variable xi represented by a leading column

ExampleConsider the following matrix:

(

0 5 70 0 0

)

Row 1 is a leading row with leading entry 5.Row 2 is a non-leading row.Column 2 is a leading column but columns 1 and 3 are not.Suppose that x1, x2, x3 are the variables represented by the columns.Then x2 is a leading variable whereas x1 and x3 are not.

TheoremIf (A|b) has a row-echelon form (U |y), then the system of equations has

0 solutions if y is a leading column

1 solution if each variable is leading

∞ solutions otherwise.

TheoremIf (A|b) has a row-echelon form (U |y), then the system of equations has

0 solutions if y is a leading column

1 solution if each variable is leading

∞ solutions otherwise.

Note that

there is a unique solution if each column is leading.

there are infinitely many solutions ify is not a leading column and at least one variable is non-leading;these variables are the parameter variables for the solutions.

TheoremIf (A|b) has a row-echelon form (U |y), then the system of equations has

0 solutions if y is a leading column

1 solution if each variable is leading

∞ solutions otherwise.

Note that

there is a unique solution if each column is leading.

there are infinitely many solutions ify is not a leading column and at least one variable is non-leading;these variables are the parameter variables for the solutions.

ExampleSuppose that

(U |y) =

(

1 −2 10 0 2

)

.

TheoremIf (A|b) has a row-echelon form (U |y), then the system of equations has

0 solutions if y is a leading column

1 solution if each variable is leading

∞ solutions otherwise.

Note that

there is a unique solution if each column is leading.

there are infinitely many solutions ify is not a leading column and at least one variable is non-leading;these variables are the parameter variables for the solutions.

ExampleSuppose that

(U |y) =

(

1 −2 10 0 2

)

.

Here, y is leading, and there is no solution since 0x1 + 0x2 6= 2.

TheoremIf (A|b) has a row-echelon form (U |y), then the system of equations has

0 solutions if y is a leading column

1 solution if each variable is leading

∞ solutions otherwise.

Note that

there is a unique solution if each column is leading.

there are infinitely many solutions ify is not a leading column and at least one variable is non-leading;these variables are the parameter variables for the solutions.

ExampleSuppose that

(U |y) =

(

1 −2 10 2 2

)

.

TheoremIf (A|b) has a row-echelon form (U |y), then the system of equations has

0 solutions if y is a leading column

1 solution if each variable is leading

∞ solutions otherwise.

ExampleSuppose that

(U |y) =

(

1 −2 10 2 2

)

.

TheoremIf (A|b) has a row-echelon form (U |y), then the system of equations has

0 solutions if y is a leading column

1 solution if each variable is leading

∞ solutions otherwise.

ExampleSuppose that

(U |y) =

(

1 −2 10 2 2

)

.

Here, both variables are leading (as are both rows and both columns of U).Row-reducing gives

(U ′|y′) =

(

1 0 30 1 1

)

so (x1, x2) = (3, 1) is the unique solution.

TheoremIf (A|b) has a row-echelon form (U |y), then the system of equations has

0 solutions if y is a leading column

1 solution if each variable is leading

∞ solutions otherwise.

Note that

there is a unique solution if each column is leading.

there are infinitely many solutions ify is not a leading column and at least one variable is non-leading;these variables are the parameter variables for the solutions.

Example

Suppose that (U |y) =

(

1 −2 10 0 0

)

.

TheoremIf (A|b) has a row-echelon form (U |y), then the system of equations has

0 solutions if y is a leading column

1 solution if each variable is leading

∞ solutions otherwise.

Note that

there is a unique solution if each column is leading.

there are infinitely many solutions ify is not a leading column and at least one variable is non-leading;these variables are the parameter variables for the solutions.

Example

Suppose that (U |y) =

(

1 −2 10 0 0

)

.

Here, x2 and y are non-leading, and the infinitely many solutions are

x =

(

x1

x2

)

=

(

1 + 2λλ

)

=

(

10

)

+ λ

(

21

)

, λ ∈ R .

TheoremLet A be an m×n matrix with a row-echelon form U . Then Ax = b has

at least 1 solution for each b∈Rn ⇔ all rows in U are leading

at most 1 solution for each b∈Rn ⇔ all columns in U are leading

exactly 1 solution for each b∈Rn ⇔ all rows/columns in U are leading

TheoremLet A be an m×n matrix with a row-echelon form U . Then Ax = b has

at least 1 solution for each b∈Rn ⇔ all rows in U are leading

at most 1 solution for each b∈Rn ⇔ all columns in U are leading

exactly 1 solution for each b∈Rn ⇔ all rows/columns in U are leading

Note that if Ax = b has

at least 1 solution for each b ∈ Rn, then n ≥ m.

at most 1 solution for each b ∈ Rn, then m ≥ n.

exactly 1 solution for each b ∈ Rn, then m = n.

TheoremLet A be an m×n matrix with a row-echelon form U . Then Ax = b has

at least 1 solution for each b∈Rn ⇔ all rows in U are leading

at most 1 solution for each b∈Rn ⇔ all columns in U are leading

exactly 1 solution for each b∈Rn ⇔ all rows/columns in U are leading

Example

Suppose that U =

(

1 −2 10 2 2

)

.

TheoremLet A be an m×n matrix with a row-echelon form U . Then Ax = b has

at least 1 solution for each b∈Rn ⇔ all rows in U are leading

at most 1 solution for each b∈Rn ⇔ all columns in U are leading

exactly 1 solution for each b∈Rn ⇔ all rows/columns in U are leading

Example

Suppose that U =

(

1 −2 10 2 2

)

.

Here, both rows of U are leading and column 3 is non-leading.

TheoremLet A be an m×n matrix with a row-echelon form U . Then Ax = b has

at least 1 solution for each b∈Rn ⇔ all rows in U are leading

at most 1 solution for each b∈Rn ⇔ all columns in U are leading

exactly 1 solution for each b∈Rn ⇔ all rows/columns in U are leading

Example

Suppose that U =

(

1 −2 10 2 2

)

.

Here, both rows of U are leading and column 3 is non-leading.For each b ∈ R

n, (A|b) can be transformed into (U |y) for some y ∈ R2

and further reduced to

(U ′|y′) =

(

1 0 3 y′1

0 1 1 y′2

)

.

TheoremLet A be an m×n matrix with a row-echelon form U . Then Ax = b has

at least 1 solution for each b∈Rn ⇔ all rows in U are leading

at most 1 solution for each b∈Rn ⇔ all columns in U are leading

exactly 1 solution for each b∈Rn ⇔ all rows/columns in U are leading

Example

Suppose that U =

(

1 −2 10 2 2

)

.

Here, both rows of U are leading and column 3 is non-leading.For each b ∈ R

n, (A|b) can be transformed into (U |y) for some y ∈ R2

and further reduced to

(U ′|y′) =

(

1 0 3 y′1

0 1 1 y′2

)

.

The system Ax = b thus has the infinite solutions

x =

(

x1

x2

)

=

(

y′1− 3λ

y′2− λ

)

=

(

y′1

y′2

)

+ λ

(

−3−1

)

, λ ∈ R .

TheoremLet A be an m×n matrix with a row-echelon form U . Then Ax = b has

at least 1 solution for each b∈Rn ⇔ all rows in U are leading

at most 1 solution for each b∈Rn ⇔ all columns in U are leading

exactly 1 solution for each b∈Rn ⇔ all rows/columns in U are leading

ExampleSuppose that

U =

1 00 10 0

.

TheoremLet A be an m×n matrix with a row-echelon form U . Then Ax = b has

at least 1 solution for each b∈Rn ⇔ all rows in U are leading

at most 1 solution for each b∈Rn ⇔ all columns in U are leading

exactly 1 solution for each b∈Rn ⇔ all rows/columns in U are leading

ExampleSuppose that

U =

1 00 10 0

.

For each b ∈ Rn, Ax = b either has a unique solution or no solution.

TheoremLet A be an m×n matrix with a row-echelon form U . Then Ax = b has

at least 1 solution for each b∈Rn ⇔ all rows in U are leading

at most 1 solution for each b∈Rn ⇔ all columns in U are leading

exactly 1 solution for each b∈Rn ⇔ all rows/columns in U are leading

ExampleSuppose that

U =

1 00 10 0

.

For each b ∈ Rn, Ax = b either has a unique solution or no solution.

ExampleSuppose that

U =

(

1 20 1

)

.

TheoremLet A be an m×n matrix with a row-echelon form U . Then Ax = b has

at least 1 solution for each b∈Rn ⇔ all rows in U are leading

at most 1 solution for each b∈Rn ⇔ all columns in U are leading

exactly 1 solution for each b∈Rn ⇔ all rows/columns in U are leading

ExampleSuppose that

U =

1 00 10 0

.

For each b ∈ Rn, Ax = b either has a unique solution or no solution.

ExampleSuppose that

U =

(

1 20 1

)

.

Here, Ax = b has a unique solution for each b ∈ Rn.

ExampleSolve the system of equations

2x1 + 3x2 = b13x1 + 4x2 = b2

ExampleSolve the system of equations

2x1 + 3x2 = b13x1 + 4x2 = b2

(

2 3 b13 4 b2

)

ExampleSolve the system of equations

2x1 + 3x2 = b13x1 + 4x2 = b2

(

2 3 b13 4 b2

)

R2−3

2R1

−−−−−→

(

2 3 b10 −1

2−3

2b1 + b2

)

ExampleSolve the system of equations

2x1 + 3x2 = b13x1 + 4x2 = b2

(

2 3 b13 4 b2

)

R2−3

2R1

−−−−−→

(

2 3 b10 −1

2−3

2b1 + b2

)

R2=−2R2−−−−−→

(

2 3 b10 1 3b1 − 2b2

)

ExampleSolve the system of equations

2x1 + 3x2 = b13x1 + 4x2 = b2

(

2 3 b13 4 b2

)

R2−3

2R1

−−−−−→

(

2 3 b10 −1

2−3

2b1 + b2

)

R2=−2R2−−−−−→

(

2 3 b10 1 3b1 − 2b2

)

R1−3R2−−−−→

(

2 0 −8b1 + 6b20 1 3b1 − 2b2

)

ExampleSolve the system of equations

2x1 + 3x2 = b13x1 + 4x2 = b2

(

2 3 b13 4 b2

)

R2−3

2R1

−−−−−→

(

2 3 b10 −1

2−3

2b1 + b2

)

R2=−2R2−−−−−→

(

2 3 b10 1 3b1 − 2b2

)

R1−3R2−−−−→

(

2 0 −8b1 + 6b20 1 3b1 − 2b2

)

R1=1

2R1

−−−−−→

(

1 0 −4b1 + 3b20 1 3b1 − 2b2

)

ExampleSolve the system of equations

2x1 + 3x2 = b13x1 + 4x2 = b2

(

2 3 b13 4 b2

)

R2−3

2R1

−−−−−→

(

2 3 b10 −1

2−3

2b1 + b2

)

R2=−2R2−−−−−→

(

2 3 b10 1 3b1 − 2b2

)

R1−3R2−−−−→

(

2 0 −8b1 + 6b20 1 3b1 − 2b2

)

R1=1

2R1

−−−−−→

(

1 0 −4b1 + 3b20 1 3b1 − 2b2

)

There is thus a unique solution:

x1 = −4b1 + 3b2x2 = 3b1 − 2b2

ExampleFind conditions on b1, b2 for solutions to exist for the system

2x1 + 3x2 = b14x1 + 6x2 = b2

Find all solutions when they exist.

ExampleFind conditions on b1, b2 for solutions to exist for the system

2x1 + 3x2 = b14x1 + 6x2 = b2

Find all solutions when they exist.

We have (

2 3 b14 6 b2

)

ExampleFind conditions on b1, b2 for solutions to exist for the system

2x1 + 3x2 = b14x1 + 6x2 = b2

Find all solutions when they exist.

We have (

2 3 b14 6 b2

)

R2−2R1−−−−→

(

2 3 b10 0 b2 − 2b1

)

ExampleFind conditions on b1, b2 for solutions to exist for the system

2x1 + 3x2 = b14x1 + 6x2 = b2

Find all solutions when they exist.

We have (

2 3 b14 6 b2

)

R2−2R1−−−−→

(

2 3 b10 0 b2 − 2b1

)

For solutions to exist, we must have b2 − 2b1 = 0.

ExampleFind conditions on b1, b2 for solutions to exist for the system

2x1 + 3x2 = b14x1 + 6x2 = b2

Find all solutions when they exist.

We have (

2 3 b14 6 b2

)

R2−2R1−−−−→

(

2 3 b10 0 b2 − 2b1

)

For solutions to exist, we must have b2 − 2b1 = 0.When this is true, we have 2x1 + 3λ = b1, so the solutions are:

x =

(

x1

x2

)

=

(

1

2b1 −

3

2λ

λ

)

=

(

1

2b10

)

+ λ

(

−3

2

1

)

, λ ∈ R

ExerciseFind conditions on b1, b2, b3, b4 for solutions to exist for the system

x1 − x2 = b12x1 + x2 = b23x1 − 3x2 = b34x1 − x2 = b4

Find all solutions when they exist.

ExerciseFind conditions on b1, b2, b3, b4 for solutions to exist for the system

x1 − x2 = b12x1 + x2 = b23x1 − 3x2 = b34x1 − x2 = b4

Find all solutions when they exist.

1 −1 b12 1 b23 −3 b34 −1 b4

R4−4R1−−−−→

1 −1 b12 1 b23 −3 b30 3 b4 − 4b1

R3−3R1−−−−→

1 −1 b12 1 b20 0 b3 − 3b10 3 b4 − 4b1

R2−2R1−−−−→

1 −1 b10 3 b2 − 2b10 0 b3 − 3b10 3 b4 − 4b1

R4−R2−−−−→

1 −1 b10 3 b2 − 2b10 0 b3 − 3b10 0 b4 − b2 − 2b1

For solutions to exist, we must have b4 − b2 − 2b1 = 0 and b3 − 3b1 = 0.When this is true, the unique solution is (x1, x2) = (1

3(b1+b2),

1

3(b2−2b1)).