Lesson 13: Rank and Solutions to Systems of Linear Equations

Lesson 12 (Sections 14.2–3)Rank and Solutions to Systems

Math 20

October 19, 2007

Announcements

I Midterm not graded yet.

I Problem Set 5 is on the WS. Due October 24

I OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)

I Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)

Summary of Last time

The linear independence of a set measures its redundancy.

Deciding linear dependence

We showed

a1, . . . , an LD ⇐⇒ c1a1 + · · ·+ cnan = 0 has a nonzero sol’n

⇐⇒(a1 . . . an

)︸︷︷︸A

c1...

cn

︸︷︷︸

c

= 0 has a nonzero sol’n

⇐⇒ system has some free variables

⇐⇒ rref(A) has a column with no leading entry to it

Deciding linear independence

So

a1, . . . , an LI ⇐⇒ every column of rref(A) has a leading entry to it

⇐⇒ A ∼(

InO

)

Relation to invertibility

Let A be an n × n matrix. If A has an inverse A−1, the onlysolution to Ac = 0 is the zero solution.

This means that there is no linear dependence relation among thecolumns.

FactA is invertible if and only if the columns of A are linearlyindependent, if and only if rref(A) = I.


Let A be an n × n matrix. If A has an inverse A−1, the onlysolution to Ac = 0 is the zero solution.This means that there is no linear dependence relation among thecolumns.




FactA is invertible if and only if the columns of A are linearlyindependent,

if and only if rref(A) = I.




Worksheet

Example

Solve

x+2y+z =1

2x+2y =1

x+3y+z =1

SolutionSince 1 2 1 1

1 1 0 01 3 1 1

1 0 0 1/2

0 1 0 00 0 1 1/2

we have x = 1/2, y = 0, z = 1/2.

Example

Solve

x+2y+z =1

2x+2y =1

x+3y+z =1

SolutionSince 1 2 1 1

1 1 0 01 3 1 1

1 0 0 1/2

0 1 0 00 0 1 1/2

we have x = 1/2, y = 0, z = 1/2.

Example

Solve

x+2y− z =1

2x+2y =1

x+3y−2z =1

SolutionSince 1 2 −1 1

1 2 0 01 3 −2 1

1 0 1 0

0 1 −1 00 0 0 1

we have no solution.

Example

Solve

x+2y− z =1

2x+2y =1

x+3y−2z =1


1 2 0 01 3 −2 1

1 0 1 0

0 1 −1 00 0 0 1

we have no solution.

Example

Solve

x+2y− z =3

2x+2y =4

x+3y−2z =4


1 2 0 41 3 −2 4

1 0 1 1

0 1 −1 10 0 0 0

The system is equivalent to x = 1− z, y = 1 + z, where z is free.

Example

Solve

x+2y− z =3

2x+2y =4

x+3y−2z =4


1 2 0 41 3 −2 4

1 0 1 1

0 1 −1 10 0 0 0

The system is equivalent to x = 1− z, y = 1 + z, where z is free.

Example

Solve

x+2y−3z =1

2x+4y−6z =1

3+6y−9z =1


2 4 −6 13 6 −9 1

1 2 −3 0

0 0 0 10 0 0 0

there is no solution.

Example

Solve

x+2y−3z =1

2x+4y−6z =1

3+6y−9z =1


2 4 −6 13 6 −9 1

1 2 −3 0

0 0 0 10 0 0 0

there is no solution.

The rank

DefinitionThe rank of a matrix A, written r(A) is the maximum number oflinearly independent column vectors in A.

If A is a zero matrix, wesay r(A) = 0.

The rank

DefinitionThe rank of a matrix A, written r(A) is the maximum number oflinearly independent column vectors in A. If A is a zero matrix, wesay r(A) = 0.

Computing the rank by Gaussian Elimination

FactIf A and B are row equivalent (we can get from one to another byrow operations), then r(A) = r(B).

So the rank of a matrix is equal to the rank of its RREF, which iseasy to calculate.

Computing the rank by Gaussian Elimination

FactIf A and B are row equivalent (we can get from one to another byrow operations), then r(A) = r(B).

So the rank of a matrix is equal to the rank of its RREF, which iseasy to calculate.

Example

Compute the ranks of the matrices1 2 12 2 11 3 1

1 2 −12 2 01 3 −2

1 2 −32 4 −63 6 −9

Answer.3, 2, and 1.

Example

Compute the ranks of the matrices1 2 12 2 11 3 1

1 2 −12 2 01 3 −2

1 2 −32 4 −63 6 −9

Answer.3, 2, and 1.

Computing the rank by minors

FactThe rank r(A) of a matrix is equal to the order of the largestminor of A which has nonzero determinant.

This is not an obvious fact, nor is it easy to prove.

Computing the rank by minors

FactThe rank r(A) of a matrix is equal to the order of the largestminor of A which has nonzero determinant.

This is not an obvious fact, nor is it easy to prove.

Rank and consistency

FactLet A be an m× n matrix, b an n× 1 vector, and Ab the matrix Aaugmented by b.

Then the system of linear equations Ax = b has a solution (isconsistent) if and only if r(A) = r(Ab).

Rank and consistency

FactLet A be an m× n matrix, b an n× 1 vector, and Ab the matrix Aaugmented by b.Then the system of linear equations Ax = b has a solution (isconsistent) if and only if r(A) = r(Ab).

Rank and redundancy

FactLet A be an m× n matrix, b an n× 1 vector, and Ab the matrix Aaugmented by b. Suppose that r(A) = r(Ab) = k < m (m is thenumber of equations in the system Ax = b).

Then m − k of the equations are redundant; they can be removedand the system has the same solutions.

Rank and redundancy

FactLet A be an m× n matrix, b an n× 1 vector, and Ab the matrix Aaugmented by b. Suppose that r(A) = r(Ab) = k < m (m is thenumber of equations in the system Ax = b).Then m − k of the equations are redundant; they can be removedand the system has the same solutions.

Rank and redundancy

FactLet A be an m× n matrix, b an n× 1 vector, and Ab the matrix Aaugmented by b. Suppose that r(A) = r(Ab) = k < n (n is thenumber of variables in the system Ax = b).

Then n− k of the variables are free; they can be chosen at will andthe rest of the variables depend on them, getting infinitely manysolutions.

Rank and redundancy

FactLet A be an m× n matrix, b an n× 1 vector, and Ab the matrix Aaugmented by b. Suppose that r(A) = r(Ab) = k < n (n is thenumber of variables in the system Ax = b).Then n− k of the variables are free; they can be chosen at will andthe rest of the variables depend on them, getting infinitely manysolutions.

Technology

Lesson 13: Rank and Solutions to Systems of Linear Equations