1 systems of linear equations -...
TRANSCRIPT
EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
Professor Joana Amorim, [email protected]
What is on today
1 Systems of Linear Equations 11.1 Matrix notation for systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Row reduction and echelon forms 52.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Using row reduction to solve a linear system . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3 Linear Combinations and vector equations 8
4 More about Ax = b 11
5 Solution sets of Linear Systems 135.1 Homogeneous linear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135.2 Non-Homogeneous linear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1 Systems of Linear Equations
In the book:
• Section 1.1, all
We probably all have solved linear systems in 2 variables:{ax1 + bx2 = cdx1 + ex2 = f
If instead of 2 variables we have more the idea is the samea1x1 + . . . + anxn = c1b1x1 + . . . + bnxn = c2...
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
There are also 3 possibilities
• The system is consistent
– one solution
– infinite number of solutions
• The system is inconsistent- it has no solutions.
Example 1 (Similar to Ex 2, section 1.1) Solve the following system using elementary row operations.{3x1 + 6x2 = −35x1 + 7x2 = 10
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
1.1 Matrix notation for systems{3x1 + 6x2 = −35x1 + 7x2 = 10
Elementary row operations (all that we need to solve systems!!):
1. (Replacement) Replace one row by the sum of itself and the multiple of another.
2. (Interchange) Interchange two rows.
3. (Scaling) Multiply all entries in row by a nonzero constant.
• We say two matrices are row equivalent if you can get one from the other using elementary rowoperations.
• Elementary row operations are reversible!
• If you have two augmented matrices that are row equivalent, then the two systems have the samesolution set.
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
Example 2 (Exercises 11 and 13, section 1.1) Solve the following systems using matrix notationx2 + 5x3 = −4
x1 + 4x2 + 3x3 = −22x1 + 7x2 + x3 = −2
x1 − 3x3 = 8
2x1 + 2x2 + 9x3 = 7x2 + 5x3 = −2
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
2 Row reduction and echelon forms
In the book:
• Section 1.2, all
2.1 Definitions
Definition 3 (Echelon form and reduced echelon form)
Example 4
2 −3 2 13 1 −4 80 0 0 5/2
1 0 0 290 1 0 160 0 1 3
0 1 0 161 0 0 290 0 1 3
1 0 1 00 1 1 00 0 0 1
1 1 0 1 10 2 0 2 20 0 0 3 30 0 0 0 4
• Echelon form not unique: all non zero matrices can be row reduced to more than one echelonmatrix;
• Reduced echelon form is unique: all non zero matrices can be reduced to one and only onereduced echelon matrix.
• Pivot: is a number corresponding to a leading 1 in the reduced echelon form of a matrix.
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
2.2 Using row reduction to solve a linear system
Remark 5 If the row echelon form of the matrix has a line of the form[0 . . . 0 b
]where b 6= 0 the
system is inconsistent.
Example 6 (Similar to Exs 9, 12, 7 and 14 section 1.2) Find the general solutions of the systemswhose augmented matrices are given.[
0 1 −2 31 −3 4 6
]
1 0 −9 0 40 1 3 0 −10 0 0 1 −70 0 0 0 1
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
[1 3 4 73 9 7 6
]
1 0 −5 0 −8 30 1 4 −1 0 60 0 0 0 1 00 0 0 0 0 0
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
3 Linear Combinations and vector equations
In the book:
• Section 1.3, pages 28 – 32.
• You can now do all exercises from section 1.3
If we havev1,v2, . . . ,vp vectors in Rn,
c1, c2, . . . , cp scalars,
then a Linear Combination of these vectors with weights c1, c2, . . . , cp is
c1v1 + c2v2 + . . . + cpvp.
Geometrically
Take v1 =
[11
]and v2 =
[−12
].
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
Some more notation...
• The set of all linear combinations of a vector v is the set spanned by v =span{v} = {cv for all scalars c}.
• The set of all linear combinations of the vectors v1 and v2 is the set spanned by v1 and v2
=span{v1,v2} = {c1v1 + c2v2 for all scalars c1, c2}.
• . . .
Important Question!: How do we decide if a given vector is the span of a set of vectors??
Example 7 Is
[03
]in span
{v1 =
[11
],v2 =
[−12
]}?
This is the same as asking: are there constants x1, x2 such that
x1v1 + x2v2 =
[03
]?
To decide if a vector
a1...an
is in the span of a set of vectors v1,v2, . . . ,vp we need to solve the
vector equation
c1v1 + c2v2 + . . . + cpvp =
a1...an
.
This is equivalent to solve the system
[v1 v2 . . . vp]
c1...cp
=
a1...an
• If the system is consistent the answer is yes- the solution (or one of the solutions) give the
coefficients;
• If the system is inconsistent the answer is no.
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
Example 8 (Exercise 13, section 1.3) Determine if b is a linear combination of the columns of thematrix A.
A =
1 −4 20 3 5−2 8 −4
, b =
3−7−3
Example 9 (Similiar to Ex 17, section 1.3)
a1 =
13−1
a2 =
−5−82
b =
3−5h
For what values of h is b in span{a1, a2}?
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
Example 10 (Similar to Ex 27, section 1.3)
4 More about Ax = b
In the book:
• Section 1.4, pages 37 – 38.
• You can now do all exercises from section 1.4
Theorem 11
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
Example 12 (Exercise 13, section 1.4) Let u =
044
and A =
3 −5−2 61 1
. Is u in the plan in R3
spanned by the columns of A? If yes write u as a linear combination of the columns of A.
Example 13 (Similar to Ex 15, section 1.4) Let A =
[3 −1−9 3
]and b =
[b1b2
]. Show that the equa-
tion Ax = b does not have a solution for all possible b, and describe the set of all b for which Ax = bdoes have a solution.
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
5 Solution sets of Linear Systems
In the book:
• All section 1.5
We want now look at all the vectors that are solution of a given system- the solution set- and a new wayto represent it using vectors.
5.1 Homogeneous linear systems
A linear system is homogeneous if it is of the form
Ax = 0.
• The zero vector is always a solution of the system Ax = 0. It is called the trivial solution.
• The interesting question is: when does the system Ax = 0 have a non-trivial solution, i.e., a solutionthat is not the zero vector?
Example 14 Give the solution of the given homogeneous system in parametric vector form (vectorsmultiplied by some real parameters):
x1 − 3x2 − 8x3 + 5x4 = 0x2 + 2x3 − 4x4 = 0
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
Example 15 (Similar to Ex 9, section 1.5) Describe all solutions of the system Ax = 0 in vectorparametric form, where A is the matrix [
3 −6 6−2 4 −2
].
5.2 Non-Homogeneous linear systems
A non-homogeneous system is a system of the form
Ax = b
where b is not the zero vector.
When a non-homogeneous system has an infinite number of solutions we can write its set of solutionsin parametric vector form as
one vector plus an arbitrary linear combination of vectors that verify the corresponding homogeneoussystem.
Theorem 16
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
Example 17 (Similar to Ex 15, section 1.5) Describe the solution set of the following system in para-metric form:
x1 + 2x2 − 3x3 = 52x1 + 1x2 − 3x3 = 13
−x1 + x2 = −8
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EK102: Linear Algebra Lectures 3, 4: Jan 30th, Feb 1st 2017 Section C1
Example 18 (Mid term, fall 2016 section B1) Let A =
1 2 −12 4 −23 6 2
and b =
36−1
. Solve, if pos-
sible, the system of equations given by the matrix equation Ax = b. If the system is consistent present thesolution in vector parametric form and
• identify a vector which is solution of the corresponding homogeneous system and a particular solution,
• describe geometrically the solution.
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