Fall 2017

Solution sets

Authors:Alexander Knop

Institute:UC San Diego

Computation of AxEXAMPLE

Let us compute Ax where

A =

2 3 4−1 5 −36 −2 8

and x =

x1x2x3

By the definition 2 3 4−1 5 −36 −2 8

x1x2x3

= x1

2−16

+ x2

35−2

x3

4−38

=

2x1 + 3x2 + 4x3−x1 + 5x2 − 3x36x1 − 2x2 + 8x3

Alexander Knop 2

Computation of AxEXAMPLE

Let us compute Ax where

A =

2 3 4−1 5 −36 −2 8

and x =

x1x2x3

By the definition 2 3 4

−1 5 −36 −2 8

x1x2x3

= x1

2−16

+ x2

35−2

x3

4−38

=

2x1 + 3x2 + 4x3−x1 + 5x2 − 3x36x1 − 2x2 + 8x3

Alexander Knop 2

Computation of Ax

ROW-VECTOR RULE FOR COMPUTING AX

If the product Ax is defined, then the ith entry in Ax is the sum of theproducts of corresponding entries from row i of A and from the vector x.

EXAMPLE

Let us compute

[1 2 30 5 −1

] 12−1

=

[1 · 1 + 2 · 2 + 3 · (−1)

0 · 1 + 5 · 2 + (−1) · (−1)

]=

[211

]

Alexander Knop 3

Computation of Ax

ROW-VECTOR RULE FOR COMPUTING AX

If the product Ax is defined, then the ith entry in Ax is the sum of theproducts of corresponding entries from row i of A and from the vector x.

EXAMPLE

Let us compute

[1 2 30 5 −1

] 12−1

=

[1 · 1 + 2 · 2 + 3 · (−1)

0 · 1 + 5 · 2 + (−1) · (−1)

]=

[211

]

Alexander Knop 3

Computation of Ax

ROW-VECTOR RULE FOR COMPUTING AX

If the product Ax is defined, then the ith entry in Ax is the sum of theproducts of corresponding entries from row i of A and from the vector x.

EXAMPLE

Let us compute

[1 2 30 5 −1

] 12−1

=

[1 · 1 + 2 · 2 + 3 · (−1)

0 · 1 + 5 · 2 + (−1) · (−1)

]

=

[211

]

Alexander Knop 3

Computation of Ax

ROW-VECTOR RULE FOR COMPUTING AX

If the product Ax is defined, then the ith entry in Ax is the sum of theproducts of corresponding entries from row i of A and from the vector x.

EXAMPLE

Let us compute

[1 2 30 5 −1

] 12−1

=

[1 · 1 + 2 · 2 + 3 · (−1)

0 · 1 + 5 · 2 + (−1) · (−1)

]=

[211

]

Alexander Knop 3

Properties of Matrix-Vector Product

THEOREM

If A is an m × n matrix, u, v ∈ Rn, and c is a scalar, then:1 A(u + v) = Au + Av;

2 A(cu) = c(Au).

Alexander Knop 4

Properties of Matrix-Vector Product

PROOF.

Let us consider case when n = 3.

A(u + v) =[a1 a2 a3

] u1 + v1u2 + v2u3 + v3

=

(u1 + v1)a1 + (u2 + v2)a2 + (u3 + v3)a3 =

(u1a1 + u2a2 + u3a3) + (v1a1 + v2a2 + v3a3) = uA + vA

Alexander Knop 4

Properties of Matrix-Vector Product

THEOREM

If A is an m × n matrix, u, v ∈ Rn, and c is a scalar, then:1 A(u + v) = Au + Av;

2 A(cu) = c(Au).

Alexander Knop 4

Properties of Matrix-Vector Product

PROOF.

Let us consider case when n = 3.

A(cu) =[a1 a2 a3

] cu1

cu2

cu3

=

(cu1)a1 + (cu2)a2 + (cu3)a3 =

c(u1a1) + c(u2a2) + c(u3a3) = c(Au)

Alexander Knop 4

Homogeneous Linear Systems

DEFINITION

The system of linear equations is called homogeneous iff it can berewritten in the form Ax = 0.Solution of homogeneous system is nontrivial iff it is not equal to zerovector.

Alexander Knop 5

Homogeneous Linear Systems

DEFINITION

The system of linear equations is called homogeneous iff it can berewritten in the form Ax = 0.Solution of homogeneous system is nontrivial iff it is not equal to zerovector.

EXAMPLE

The solution (1, 2, 1) of the system

x1 + x2 + x3 = 4

x1 − x3 = 0is nontrivial.Alexander Knop 5

Homogeneous Linear Systems

DEFINITION

The system of linear equations is called homogeneous iff it can berewritten in the form Ax = 0.Solution of homogeneous system is nontrivial iff it is not equal to zerovector.

THEOREM

The homogeneous system has nontrivial solution iff the system has atleast one free variable.

Alexander Knop 5

Homogeneous Linear Systems

EXAMPLE

Let us determine if the following system has nontrivial solution.

3x1 + 5x2 − 4x3 = 0

−3x1 − 2x2 + 4x3 = 0

6x1 + x2 − 8x3 = 0

Alexander Knop 6

Homogeneous Linear Systems

EXAMPLE

Let us determine if the following system has nontrivial solution.

3x1 + 5x2 − 4x3 = 0

−3x1 − 2x2 + 4x3 = 0

6x1 + x2 − 8x3 = 0

Let us consider the augmented matrix. 3 5 −4 0−3 −2 4 06 1 −8 0

∼

3 5 −4 00 3 0 00 −9 0 0

∼

3 5 −4 00 3 0 00 0 0 0

Alexander Knop 6

Homogeneous Linear Systems

EXAMPLE

Let us determine if the following system has nontrivial solution.

3x1 + 5x2 − 4x3 = 0

−3x1 − 2x2 + 4x3 = 0

6x1 + x2 − 8x3 = 0

In the reduced echelon form this system equals1 0 − 43 0

0 1 0 00 0 0 0

x1 − 43x3 = 0

x2 = 0

+ 0 = 0

Alexander Knop 6

Homogeneous Linear Systems

EXAMPLE

Let us determine if the following system has nontrivial solution.

3x1 + 5x2 − 4x3 = 0

−3x1 − 2x2 + 4x3 = 0

6x1 + x2 − 8x3 = 0

Solution of this system is

x =

43x30x3

Alexander Knop 6

Homogeneous Linear Systems

EXAMPLE

Let us determine if the following system has nontrivial solution.

3x1 + 5x2 − 4x3 = 0

−3x1 − 2x2 + 4x3 = 0

6x1 + x2 − 8x3 = 0

Solution of this system is

x = x3

4301

= x3v

Alexander Knop 6

Nonhomogeneous Linear Systems

EXAMPLE

Let us solve the following system.

3x1 + 5x2 − 4x3 = 7

−3x1 − 2x2 + 4x3 = −1

6x1 + x2 − 8x3 = −4

Alexander Knop 7

Nonhomogeneous Linear Systems

EXAMPLE

Let us solve the following system.

3x1 + 5x2 − 4x3 = 7

−3x1 − 2x2 + 4x3 = −1

6x1 + x2 − 8x3 = −4

Using row operations we obtain reduced echelon form.1 0 −43 −1

0 1 0 20 0 0 0

x1 − 43x3 = −1

x2 = 2

+ 0 = 0

Alexander Knop 7

Nonhomogeneous Linear Systems

EXAMPLE

Let us solve the following system.

3x1 + 5x2 − 4x3 = 7

−3x1 − 2x2 + 4x3 = −1

6x1 + x2 − 8x3 = −4

As a vector general solution looks like:

x =

−1 + 43x3

2x3

=

−120

+

43x30x3

=

−120

+ x3

4301

= p + x3v

Alexander Knop 7

Structure of Solution Sets

THEOREM

Let Ax = b be a consistent linear system and and let p be a solution.Then the solution set of Ax = b is the set of all vectors of the formw = p + vh where is any solution of the homogeneous system Ax = 0.

Alexander Knop 8

Structure of Solution Sets

WRITING A SOLUTION SET IN PARAMETRIC VECTOR FORM

1 Row reduce augmented matrix to reduced echelon form.

2 Express each basic variable in terms of any free variables appearingin an equation.

3 Write a typical solution x as vector entries depends on freevariables, if any.

4 Decompose x into linear combination of vectors (with numericentries) using the free variables as parameters.

Alexander Knop 9

Structure of Solution Sets

WRITING A SOLUTION SET IN PARAMETRIC VECTOR FORM

1 Row reduce augmented matrix to reduced echelon form.

2 Express each basic variable in terms of any free variables appearingin an equation.

3 Write a typical solution x as vector entries depends on freevariables, if any.

4 Decompose x into linear combination of vectors (with numericentries) using the free variables as parameters.

Alexander Knop 9

Structure of Solution Sets

WRITING A SOLUTION SET IN PARAMETRIC VECTOR FORM

1 Row reduce augmented matrix to reduced echelon form.

2 Express each basic variable in terms of any free variables appearingin an equation.

3 Write a typical solution x as vector entries depends on freevariables, if any.

4 Decompose x into linear combination of vectors (with numericentries) using the free variables as parameters.

Alexander Knop 9

Structure of Solution Sets

WRITING A SOLUTION SET IN PARAMETRIC VECTOR FORM

1 Row reduce augmented matrix to reduced echelon form.

2 Express each basic variable in terms of any free variables appearingin an equation.

3 Write a typical solution x as vector entries depends on freevariables, if any.

4 Decompose x into linear combination of vectors (with numericentries) using the free variables as parameters.

Alexander Knop 9

Structure of Solution Sets

EXAMPLE

Let us consider a matrix [1 2 1 23 3 0 3

]

Transform it into reduced echelon form[1 2 1 20 −3 −3 3

]∼

[1 2 1 20 1 1 −1

]∼

[1 0 −1 40 1 1 1

]x1

x2x3

=

4 + x31− x3

x3

= x3

1−11

+

410

Alexander Knop 10

Structure of Solution Sets

EXAMPLE

Let us consider a matrix [1 2 1 23 3 0 3

]Transform it into reduced echelon form[

1 2 1 20 −3 −3 3

]

∼[1 2 1 20 1 1 −1

]∼

[1 0 −1 40 1 1 1

]x1

x2x3

=

4 + x31− x3

x3

= x3

1−11

+

410

Alexander Knop 10

Structure of Solution Sets

EXAMPLE

Let us consider a matrix [1 2 1 23 3 0 3

]Transform it into reduced echelon form[

1 2 1 20 −3 −3 3

]∼

[1 2 1 20 1 1 −1

]

∼[1 0 −1 40 1 1 1

]x1

x2x3

=

4 + x31− x3

x3

= x3

1−11

+

410

Alexander Knop 10

Structure of Solution Sets

EXAMPLE

Let us consider a matrix [1 2 1 23 3 0 3

]Transform it into reduced echelon form[

1 2 1 20 −3 −3 3

]∼

[1 2 1 20 1 1 −1

]∼

[1 0 −1 40 1 1 1

]

x1x2x3

=

4 + x31− x3

x3

= x3

1−11

+

410

Alexander Knop 10

Structure of Solution Sets

EXAMPLE

Let us consider a matrix [1 2 1 23 3 0 3

]Transform it into reduced echelon form[

1 2 1 20 −3 −3 3

]∼

[1 2 1 20 1 1 −1

]∼

[1 0 −1 40 1 1 1

]x1

x2x3

=

4 + x31− x3

x3

= x3

1−11

+

410

Alexander Knop 10

Structure of Solution Sets

EXAMPLE

Let us consider a matrix [1 2 1 23 3 0 3

]Transform it into reduced echelon form[

1 2 1 20 −3 −3 3

]∼

[1 2 1 20 1 1 −1

]∼

[1 0 −1 40 1 1 1

]x1

x2x3

=

4 + x31− x3

x3

= x3

1−11

+

410

Alexander Knop 10