Euclidean m-Space & Linear Equations Systems of Linear Equations.
Lesson 13: Rank and Solutions to Systems of Linear Equations
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Transcript of 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.
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.
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.
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.
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
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 =3
2x+2y =4
x+3y−2z =4
SolutionSince 1 2 −1 3
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
SolutionSince 1 2 −1 3
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
SolutionSince 1 2 −3 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
SolutionSince 1 2 −3 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.