linear combination, matrix equations
TRANSCRIPT
Announcements
Ï If anyone has not been able to access the class website please
email me at [email protected]
Ï No class on Monday (Martin Luther King day)
Ï Quiz 1 in class on Wednesday Jan 20 on sections 1.1, 1.2 and
1.3
Ï Please know all de�nitions clearly for the quiz. The quiz will
not be about doing lengthy and tiresome row operations but
smaller problems to see whether you know the concepts.
Sec 1.3 Contd: Vector Equations
Vectors in n-dimensions, Rn
Ï Just like we had an ordered pair for x −y (or R2) and ordered
triplet for x −y −z (or R3), we have ordered n−tuples forn-dimensions (or Rn)
Ï Obviously not possible to draw or visualize
Ï Vectors in Rn is a n×1 column matrix of the form u=
u1u2u3...
un
Basic Algebraic Properties
For all u, v and w in Rn and all scalars c and d
Ï u+v=v+u
Ï (u+v)+w=u+(v+w)
Ï u+0=0+u=u
Ï u+(-u)=-u+u=0
Ï c(u+v)=cu+cv
Ï (c +d)u=cu+du
Ï c(du)=(cd)(u)
Ï 1u=u
Linear Combinations
Let v1, v2, v3, . . ., vp be vectors in Rn and let c1, c2, c3, . . ., cp be
scalars.
We can de�ne a new vector y as follows
y = c1v1 + c2v2 + c3v3 + . . . + cpvp
This new vector y is called a LINEAR COMBINATION of v1, v2,v3, . . ., vp with WEIGHTS c1, c2, c3, . . ., cp
Linear Combinations
Let v1, v2, v3, . . ., vp be vectors in Rn and let c1, c2, c3, . . ., cp be
scalars.
We can de�ne a new vector y as follows
y = c1v1 + c2v2 + c3v3 + . . . + cpvp
This new vector y is called a LINEAR COMBINATION of v1, v2,v3, . . ., vp with WEIGHTS c1, c2, c3, . . ., cp
Linear Combinations
Let v1, v2, v3, . . ., vp be vectors in Rn and let c1, c2, c3, . . ., cp be
scalars.
We can de�ne a new vector y as follows
y = c1v1 + c2v2 + c3v3 + . . . + cpvp
This new vector y is called a LINEAR COMBINATION of v1, v2,v3, . . ., vp with WEIGHTS c1, c2, c3, . . ., cp
Simple Examples
Let v1 and v2 be vectors in R2
The following are examples of linear combinations of v1 and v2
5v1 + v2, v1 +p13v2, πv1 - 0.5v2, 3v2(=0v1 + 3v2)
Vector Equation
An equation with vectors on both sides!!
For vectors a1 and a2 and b in R2, the equation x1a1 + x2a2 = b is
a vector equation. Here x1 and x2 are weights. This extends to R3
and Rn
The solution set of a vector equation (we want to solve for the
weights) is the same as the solution of the linear system whose
augmented matrix is formed by the vectors written as columns.
Vector Equation
An equation with vectors on both sides!!
For vectors a1 and a2 and b in R2, the equation x1a1 + x2a2 = b is
a vector equation. Here x1 and x2 are weights. This extends to R3
and Rn
The solution set of a vector equation (we want to solve for the
weights) is the same as the solution of the linear system whose
augmented matrix is formed by the vectors written as columns.
Vector Equation
An equation with vectors on both sides!!
For vectors a1 and a2 and b in R2, the equation x1a1 + x2a2 = b is
a vector equation. Here x1 and x2 are weights. This extends to R3
and Rn
The solution set of a vector equation (we want to solve for the
weights) is the same as the solution of the linear system whose
augmented matrix is formed by the vectors written as columns.
Meaning...
Suppose we are given a1=
1
−23
, a2= 5
−13−3
and b=
−381
. Can we
�nd two scalars (weights) x1 and x2 such that we can write
x1a1 + x1a2 = b? What I said earlier was..
Ï To do this, you write the 3 vectors as the 3 columns of a
matrix, the last column corresponding to b being the
augmented column.
Ï Do row reductions to solve for x1 and x2.
Ï If the system is inconsistent (remember that "0=nonzero"?),
that means we cannot �nd such weights.
Ï If the system is inconsistent, b CANNOT be written as a linear
combination of a1 and a2.
Meaning...
Suppose we are given a1=
1
−23
, a2= 5
−13−3
and b=
−381
. Can we
�nd two scalars (weights) x1 and x2 such that we can write
x1a1 + x1a2 = b? What I said earlier was..
Ï To do this, you write the 3 vectors as the 3 columns of a
matrix, the last column corresponding to b being the
augmented column.
Ï Do row reductions to solve for x1 and x2.
Ï If the system is inconsistent (remember that "0=nonzero"?),
that means we cannot �nd such weights.
Ï If the system is inconsistent, b CANNOT be written as a linear
combination of a1 and a2.
Meaning...
Suppose we are given a1=
1
−23
, a2= 5
−13−3
and b=
−381
. Can we
�nd two scalars (weights) x1 and x2 such that we can write
x1a1 + x1a2 = b? What I said earlier was..
Ï To do this, you write the 3 vectors as the 3 columns of a
matrix, the last column corresponding to b being the
augmented column.
Ï Do row reductions to solve for x1 and x2.
Ï If the system is inconsistent (remember that "0=nonzero"?),
that means we cannot �nd such weights.
Ï If the system is inconsistent, b CANNOT be written as a linear
combination of a1 and a2.
Meaning...
Suppose we are given a1=
1
−23
, a2= 5
−13−3
and b=
−381
. Can we
�nd two scalars (weights) x1 and x2 such that we can write
x1a1 + x1a2 = b? What I said earlier was..
Ï To do this, you write the 3 vectors as the 3 columns of a
matrix, the last column corresponding to b being the
augmented column.
Ï Do row reductions to solve for x1 and x2.
Ï If the system is inconsistent (remember that "0=nonzero"?),
that means we cannot �nd such weights.
Ï If the system is inconsistent, b CANNOT be written as a linear
combination of a1 and a2.
So let's do this
1 5 −3
−2 −13 8
3 −3 1
R2+2R1
and 1 5 −3
−2 −13 8
3 −3 1
R3-3R1
We get
1 5 −30 −3 2
0 −18 10
and then do
1 5 −3
0 −3 2
0 −18 10
R3-6R2
Result
1 5 −30 −3 2
0 0 −2
So the last row says 0=-2. This means the system is inconsistent.
So we CANNOT write b as a linear combination of a1 and a2.
So testing whether a vector can be written as a linear combination
of 2 or more vectors involve writing the augmented matrix and
solving for the weights.
Result
1 5 −30 −3 2
0 0 −2
So the last row says 0=-2. This means the system is inconsistent.
So we CANNOT write b as a linear combination of a1 and a2.
So testing whether a vector can be written as a linear combination
of 2 or more vectors involve writing the augmented matrix and
solving for the weights.
Result
1 5 −30 −3 2
0 0 −2
So the last row says 0=-2. This means the system is inconsistent.
So we CANNOT write b as a linear combination of a1 and a2.
So testing whether a vector can be written as a linear combination
of 2 or more vectors involve writing the augmented matrix and
solving for the weights.
Problem 14, sec 1.3
Determine if b is a linear combination of the vectors formed from
the columns of the matrix A.
A=1 −2 −60 3 7
1 −2 5
, b=11−59
Before we start, this problem is same as
Determine if b is a linear combination of a1, a2 and a3 where
a1 =
101
, a2=−23−2
, a3=−675
, b=11−59
Problem 14, sec 1.3
Determine if b is a linear combination of the vectors formed from
the columns of the matrix A.
A=1 −2 −60 3 7
1 −2 5
, b=11−59
Before we start, this problem is same as
Determine if b is a linear combination of a1, a2 and a3 where
a1 =
101
, a2=−23−2
, a3=−675
, b=11−59
Augmented matrix and row operations1 −2 −6 11
0 3 7 −5
1 −2 5 9
R3-R1
=⇒ 1 −2 −6 11
0 3 7 −50 0 11 −2
Divide R2 by 3 and R3 by 11 1 −2 −6 11
0 1 7
3−5
3
0 0 1 − 2
11
This matrix tells you that the system is consistent. That is all what
you want to know to decide whether the linear combination is
possible. Finding the complete solution here is not necessary.
Augmented matrix and row operations1 −2 −6 11
0 3 7 −5
1 −2 5 9
R3-R1
=⇒ 1 −2 −6 11
0 3 7 −50 0 11 −2
Divide R2 by 3 and R3 by 11 1 −2 −6 11
0 1 7
3−5
3
0 0 1 − 2
11
This matrix tells you that the system is consistent. That is all what
you want to know to decide whether the linear combination is
possible. Finding the complete solution here is not necessary.
Augmented matrix and row operations1 −2 −6 11
0 3 7 −5
1 −2 5 9
R3-R1
=⇒ 1 −2 −6 11
0 3 7 −50 0 11 −2
Divide R2 by 3 and R3 by 11 1 −2 −6 11
0 1 7
3−5
3
0 0 1 − 2
11
This matrix tells you that the system is consistent. That is all what
you want to know to decide whether the linear combination is
possible. Finding the complete solution here is not necessary.
Span
Span is a fancy term for a collection of linear combinations of
vectors. Here is the formal de�nition
Let v1, v2, v3, . . ., vp be vectors in Rn.
The set of all linear combinations of v1, v2, v3, . . ., vp is denoted
by Span{v1,v2, . . .vp
}
This is called the subset of Rn spanned (or generated) by v1, v2,v3, . . ., vp.
Span
Span is a fancy term for a collection of linear combinations of
vectors. Here is the formal de�nition
Let v1, v2, v3, . . ., vp be vectors in Rn.
The set of all linear combinations of v1, v2, v3, . . ., vp is denoted
by Span{v1,v2, . . .vp
}
This is called the subset of Rn spanned (or generated) by v1, v2,v3, . . ., vp.
Span
Span is a fancy term for a collection of linear combinations of
vectors. Here is the formal de�nition
Let v1, v2, v3, . . ., vp be vectors in Rn.
The set of all linear combinations of v1, v2, v3, . . ., vp is denoted
by Span{v1,v2, . . .vp
}
This is called the subset of Rn spanned (or generated) by v1, v2,v3, . . ., vp.
Span
Span{v1,v2, . . .vp
}is a collection of all vectors that can be written
as c1v1 + c2v2 + c3v3 + . . . + cpvp
If someone asks you whether a vector b is in Span {v1,v2,v3} whatshould you do?
Just do the same thing you do to check whether b is a linear
combination of v1,v2, and v3.
Or whether the linear system with augmented matrix formed by
v1,v2, v3 and b has a solution (consistent)
Span
Span{v1,v2, . . .vp
}is a collection of all vectors that can be written
as c1v1 + c2v2 + c3v3 + . . . + cpvp
If someone asks you whether a vector b is in Span {v1,v2,v3} whatshould you do?
Just do the same thing you do to check whether b is a linear
combination of v1,v2, and v3.
Or whether the linear system with augmented matrix formed by
v1,v2, v3 and b has a solution (consistent)
Span
Span{v1,v2, . . .vp
}is a collection of all vectors that can be written
as c1v1 + c2v2 + c3v3 + . . . + cpvp
If someone asks you whether a vector b is in Span {v1,v2,v3} whatshould you do?
Just do the same thing you do to check whether b is a linear
combination of v1,v2, and v3.
Or whether the linear system with augmented matrix formed by
v1,v2, v3 and b has a solution (consistent)
Problem 12 section 1.3 (Re-worded)Determine if b is in Span {a1,a2,a3} where
a1 =
1
−22
, a2=055
, a3=208
, b=−511−7
1 0 2 −5
−2 5 0 11
2 5 8 −7
R3+R2
1 0 2 −5
−2 5 0 11
0 5 4 2
R2+2R1
Result 1 0 2 −5
0 5 4 1
0 5 4 2
R3-R2
1 0 2 −5
0 5 4 1
0 0 0 1
There we have it again, 0=1!!! The system is inconsistent, no solution
which means no linear combination of b possible in terms of the 3
given vectors or b IS NOT in Span {v1,v2,v3}
Result 1 0 2 −5
0 5 4 1
0 5 4 2
R3-R2
1 0 2 −5
0 5 4 1
0 0 0 1
There we have it again, 0=1!!! The system is inconsistent, no solution
which means no linear combination of b possible in terms of the 3
given vectors or b IS NOT in Span {v1,v2,v3}
Problem 18, section 1.3
Let v1 =
1
0
−2
, v2=−318
, y= h
−5−3
. For what value(s) of h is y in
the plane generated by v1 and v2?1 −3 h
0 1 −5
−2 8 −3
R3+2R1
1 −3 h
0 1 −5
0 2 −3+2h
R3-2R2
Problem 18, section 1.3
Remember, the problem is just asking you the value(s) of h that
will make the linear system represented by this augmented matrix
consistent. 1 −3 h
0 1 −5
0 0 7+2h
To have a consistent system, just make sure that 7+2h= 0 or
h=−7
2.
Section 1.4 The Matrix Equation Ax=b
A linear combination of vectors is the product of a matrix and a
vector.
De�nitionIf A is an m×n matrix with columns formed by the vectors
a1,a2, . . .an and if x is in Rn, then the product of A and x denoted
by Ax is the linear combination of the columns of A using the
corresponding entries in x as weights.
Ax= [a1 a2 . . . an
]x1x2...
xn
= x1a1+x2a2+ . . .+xnan
Section 1.4 The Matrix Equation Ax=b
A linear combination of vectors is the product of a matrix and a
vector.
De�nitionIf A is an m×n matrix with columns formed by the vectors
a1,a2, . . .an and if x is in Rn, then the product of A and x denoted
by Ax is the linear combination of the columns of A using the
corresponding entries in x as weights.
Ax= [a1 a2 . . . an
]x1x2...
xn
= x1a1+x2a2+ . . .+xnan
Section 1.4 The Matrix Equation Ax=b
For this product to be possible, the number of columns in A mustbe same as the number of entries in x
Example
1 −3 7 1
2 1 −5 2
5 2 −3 3
1
2
3
4
= 1
125
+2
−312
+3
7
−5−3
+4
123
=125
+−624
+ 21
−15−9
+ 4
8
12
=20−312
Section 1.4 The Matrix Equation Ax=b
For this product to be possible, the number of columns in A mustbe same as the number of entries in x
Example
1 −3 7 1
2 1 −5 2
5 2 −3 3
1
2
3
4
= 1
125
+2
−312
+3
7
−5−3
+4
123
=125
+−624
+ 21
−15−9
+ 4
8
12
=20−312
Section 1.4 The Matrix Equation Ax=b
For this product to be possible, the number of columns in A mustbe same as the number of entries in x
Example
1 −3 7 1
2 1 −5 2
5 2 −3 3
1
2
3
4
= 1
125
+2
−312
+3
7
−5−3
+4
123
=125
+−624
+ 21
−15−9
+ 4
8
12
=20−312
Matrix Equation and Vector Equation
Consider the following systemx1 − 6x2 + 7x3 = 5
x2 + 2x3 = −3
This is the same as
x1
[1
0
]+x2
[−61
]+x3
[7
2
]=
[5
−3]
This is the Vector Equation.
Matrix Equation and Vector Equation
Consider the following systemx1 − 6x2 + 7x3 = 5
x2 + 2x3 = −3
This is the same as
x1
[1
0
]+x2
[−61
]+x3
[7
2
]=
[5
−3]
This is the Vector Equation.
Matrix Equation and Vector Equation
Consider the following systemx1 − 6x2 + 7x3 = 5
x2 + 2x3 = −3
This is the same as
x1
[1
0
]+x2
[−61
]+x3
[7
2
]=
[5
−3]
This is the Vector Equation.
Matrix Equation and Vector Equation
Consider the following systemx1 − 6x2 + 7x3 = 5
x2 + 2x3 = −3
This is the same as
[1 −6 7
0 1 2
]x1x2x3
=[5
−3]
This is the Matrix Equation.
Matrix Equation and Vector Equation
Consider the following systemx1 − 6x2 + 7x3 = 5
x2 + 2x3 = −3
This is the same as
[1 −6 7
0 1 2
]x1x2x3
=[5
−3]
This is the Matrix Equation.
Matrix Equation and Vector Equation
Consider the following systemx1 − 6x2 + 7x3 = 5
x2 + 2x3 = −3
This is the same as
[1 −6 7
0 1 2
]x1x2x3
=[5
−3]
This is the Matrix Equation.
Everything is the same
TheoremIf A is an m×n matrix with columns formed by the vectors
a1,a2, . . .an and if b is in Rm, then the matrix equation Ax= b has
the same solution as the following:
Ï The vector equation x1a1+x2a2+ . . .+xnan = b
Ï The system of linear equations whose augmented matrix is[a1 a2 . . . an
](which you have been all these days)
That is, solving a matrix equation is the same as solving the
corresponding vector equation which is the same as solving the
corresponding system of linear equations by row-reducing the
augmented matrix.
Everything is the same
TheoremIf A is an m×n matrix with columns formed by the vectors
a1,a2, . . .an and if b is in Rm, then the matrix equation Ax= b has
the same solution as the following:
Ï The vector equation x1a1+x2a2+ . . .+xnan = b
Ï The system of linear equations whose augmented matrix is[a1 a2 . . . an
](which you have been all these days)
That is, solving a matrix equation is the same as solving the
corresponding vector equation which is the same as solving the
corresponding system of linear equations by row-reducing the
augmented matrix.
Everything is the same
TheoremIf A is an m×n matrix with columns formed by the vectors
a1,a2, . . .an and if b is in Rm, then the matrix equation Ax= b has
the same solution as the following:
Ï The vector equation x1a1+x2a2+ . . .+xnan = b
Ï The system of linear equations whose augmented matrix is[a1 a2 . . . an
](which you have been all these days)
That is, solving a matrix equation is the same as solving the
corresponding vector equation which is the same as solving the
corresponding system of linear equations by row-reducing the
augmented matrix.
Everything is the same
TheoremIf A is an m×n matrix with columns formed by the vectors
a1,a2, . . .an and if b is in Rm, then the matrix equation Ax= b has
the same solution as the following:
Ï The vector equation x1a1+x2a2+ . . .+xnan = b
Ï The system of linear equations whose augmented matrix is[a1 a2 . . . an
](which you have been all these days)
That is, solving a matrix equation is the same as solving the
corresponding vector equation which is the same as solving the
corresponding system of linear equations by row-reducing the
augmented matrix.
Existence of Solutions
The matrix equation Ax= b has a solution if and only if b is a
linear combination of the columns of A.
So Ax= b should be consistent for all possible b for solution to exist.
TheoremIf A is an m×n matrix and if the matrix equation Ax= b has a
solution then
Ï Each b is a linear combination of the columns of A
Ï The columns of A span Rm (very important)
Ï A has pivot position in every row (not necessarily every
column). Please note that A is the coe�cient matrix (and not
the augmented matrix)
Existence of Solutions
The matrix equation Ax= b has a solution if and only if b is a
linear combination of the columns of A.
So Ax= b should be consistent for all possible b for solution to exist.
TheoremIf A is an m×n matrix and if the matrix equation Ax= b has a
solution then
Ï Each b is a linear combination of the columns of A
Ï The columns of A span Rm (very important)
Ï A has pivot position in every row (not necessarily every
column). Please note that A is the coe�cient matrix (and not
the augmented matrix)
Problem 12, section 1.4
Write the augmented matrix for the linear system corresponding to
the matrix equation Ax= b, solve the system and write the answer
as a vector.
A= 1 2 1
−3 −1 2
0 5 3
, b= 0
1
−1
The augmented matrix is
1 2 1 0
−3 −1 2 1
0 5 3 −1
R2+3R1
Problem 12, section 1.41 2 1 0
0 5 5 1
0 5 3 −1
R3-R2
1 2 1 0
0 5 5 1
0 0 −2 −2
Divide R3 by -2 1 2 1 0
0 5 5 1
0 0 1 1
Problem 12, section 1.41 2 1 0
0 5 5 1
0 0 1 1
R2-5R3
1 2 1 0
0 5 0 −40 0 1 1
Divide R2 by 5 1 2 1 0
0 1 0 −4
5
0 0 1 1
Problem 12, section 1.4
1 2 1 0
0 1 0 −4
5
0 0 1 1
R1-R3
1 2 0 −1
0 1 0 −4
5
0 0 1 1
R1-2R2
1 0 0 3
5
0 1 0 −4
5
0 0 1 1
The augmented column in the above matrix is the solution vector
x=x1x2x3
= 3
5
−4
5
1
Problem 22, section 1.4
Let v1 =
0
0
−2
, v2= 0
−38
, v3= 4
−1−5
. Does {a1,a2,a3} span R3 ?
Why or why not?
Problem 22, section 1.4
OutlineThis problem is a direct application of the theorem we saw
just now. We need to see the connection between pivots, existence
of solution and span. If we show that this matrix has pivot
in each row, we have a solution and that means these vectors span R30 0 4
0 −3 −1
−2 8 5
Swap
−2 8 5
0 −3 −1
0 0 4
Problem 22, section 1.4
Divide �rst row by -2, second row by -3 and third row by 4 1 −4 5
2
0 1 1
3
0 0 1
We have pivot position in each row which means that by our
theorem, {a1,a2,a3} span R3.
Row-Vector Rule for Computing Ax=b
If the product Ax= b is de�ned, the entry in the ith position of Axis the sum of the products of the entries from row i of A and from
the vector x (very important for matrix multiplication in chapter 2)
Example
1.
1 −3 7 1
2 1 −5 2
5 2 −3 3
1
2
3
4
= 1.1+ (−3).2+7.3+1.4
2.1+1.2+ (−5).3+2.45.1+2.2+ (−3).3+3.4
=20−312
2. 1 0 0
0 1 0
0 0 1
567
= 1.5+0.6+0.7
0.5+1.6+0.70.5+0.2+1.7
=567