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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)).