Lesson 12 (Sections 14.2–3)
Rank and Solutions to Systems
Math 20
October 19, 2007
Announcements
Midterm not graded yet.
Problem Set 5 is on the WS. Due October 24
OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)
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 ⇐⇒ c1 a1 + · · · + cn an = 0 has a nonzero sol’n

c1
.
⇐⇒ a1 . . . an  .  = 0 has a nonzero sol’n
.
cn
A
c
⇐⇒ 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
In
⇐⇒ A ∼
O

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

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

Relation to invertibility
Let A be an n × n matrix. If A has an inverse A−1 , the only
solution to Ac = 0 is the zero solution.
This means that there is no linear dependence relation among the
columns.
Fact
A is invertible if and only if the columns of A are linearly
independent,

Relation to invertibility
Let A be an n × n matrix. If A has an inverse A−1 , the only
solution to Ac = 0 is the zero solution.
This means that there is no linear dependence relation among the
columns.
Fact
A is invertible if and only if the columns of A are linearly
independent, if and only if rref(A) = I.

Worksheet

Example
Solve
x+2y +z =1
2x+2y =1
x+3y +z =1

Example
Solve
x+2y +z =1
2x+2y =1
x+3y +z =1
Solution
Since    
1211 100 1/2
1 1 0 0 0 1 0 0
1311 001 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

Example
Solve
x+2y − z =1
2x+2y =1
x+3y −2z =1
Solution
Since    
1 2 −1 1 10 10
 0 1 −1 0 
1 2 0 0
1 3 −2 1 00 01
we have no solution.

Example
Solve
x+2y − z =3
2x+2y =4
x+3y −2z =4

Example
Solve
x+2y − z =3
2x+2y =4
x+3y −2z =4
Solution
Since    
1 2 −1 3 10 11
 0 1 −1 1 
1 2 0 4
1 3 −2 4 00 00
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

Example
Solve
x+2y −3z =1
2x+4y −6z =1
3+6y −9z =1
Solution
Since    
1 2 −3 1 1 2 −3 0
 2 4 −6 1  0 0 0 1
3 6 −9 1 00 00
there is no solution.

The rank
Definition
The rank of a matrix A, written r (A) is the maximum number of
linearly independent column vectors in A.

The rank
Definition
The rank of a matrix A, written r (A) is the maximum number of
linearly independent column vectors in A. If A is a zero matrix, we
say r (A) = 0.

Computing the rank by Gaussian Elimination
Fact
If A and B are row equivalent (we can get from one to another by
row operations), then r (A) = r (B).

Computing the rank by Gaussian Elimination
Fact
If A and B are row equivalent (we can get from one to another by
row operations), then r (A) = r (B).
So the rank of a matrix is equal to the rank of its RREF, which is
easy to calculate.

Example
Compute the ranks of the matrices
     
1 2 −1 1 2 −3
1 21
2 4 −6
2 2 1 2 2 0 
1 3 −2 3 6 −9
1 31

Example
Compute the ranks of the matrices
     
1 2 −1 1 2 −3
1 21
2 4 −6
2 2 1 2 2 0 
1 3 −2 3 6 −9
1 31
Answer.
3, 2, and 1.

Computing the rank by minors
Fact
The rank r (A) of a matrix is equal to the order of the largest
minor of A which has nonzero determinant.

Computing the rank by minors
Fact
The rank r (A) of a matrix is equal to the order of the largest
minor of A which has nonzero determinant.
This is not an obvious fact, nor is it easy to prove.

Rank and consistency
Fact
Let A be an m × n matrix, b an n × 1 vector, and Ab the matrix A
augmented by b.

Rank and consistency
Fact
Let A be an m × n matrix, b an n × 1 vector, and Ab the matrix A
augmented by b.
Then the system of linear equations Ax = b has a solution (is
consistent) if and only if r (A) = r (Ab ).

Rank and redundancy
Fact
Let A be an m × n matrix, b an n × 1 vector, and Ab the matrix A
augmented by b. Suppose that r (A) = r (Ab ) = k < m (m is the
number of equations in the system Ax = b).

Rank and redundancy
Fact
Let A be an m × n matrix, b an n × 1 vector, and Ab the matrix A
augmented by b. Suppose that r (A) = r (Ab ) = k < m (m is the
number of equations in the system Ax = b).
Then m − k of the equations are redundant; they can be removed
and the system has the same solutions.

Rank and redundancy
Fact
Let A be an m × n matrix, b an n × 1 vector, and Ab the matrix A
augmented by b. Suppose that r (A) = r (Ab ) = k < n (n is the
number of variables in the system Ax = b).

Rank and redundancy
Fact
Let A be an m × n matrix, b an n × 1 vector, and Ab the matrix A
augmented by b. Suppose that r (A) = r (Ab ) = k < n (n is the
number of variables in the system Ax = b).
Then n − k of the variables are free; they can be chosen at will and
the rest of the variables depend on them, getting infinitely many
solutions.