Transcript Slide 1.4
1 Linear Equations
in Linear Algebra
1.4
THE MATRIX EQUATION Ax b
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MATRIX EQUATION Ax b
Definition: If A is an m n matrix, with columns a1,
n
…, an, and if x is in
, 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; that is,
Ax a1 a 2
x1
x
2
a n x1a1 x2a 2 ... xna n .
x
n
Ax is defined only if the number of columns of A
equals the number of entries in x.
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MATRIX EQUATION Ax b
m
Example 1: For v1, v2, v3 in
, write the linear
combination 3v1 5v 2 7v3 as a matrix times a
vector.
Solution: Place v1, v2, v3 into the columns of a matrix
A and place the weights 3, 5 , and 7 into a vector x.
That is,
3v1 5v 2 7v3 v1
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v2
3
v3 5 Ax .
7
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MATRIX EQUATION Ax b
Now, write the system of linear equations as a vector
equation involving a linear combination of vectors.
For example, the following system
x1 2 x2 x3 4
----(1)
5 x2 3x3 1
is equivalent to
1
2
1 4
x1 x2 x3 .
0
5
3 1
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----(2)
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MATRIX EQUATION Ax b
As in the given example (1), the linear combination
on the left side is a matrix times a vector, so that (2)
becomes
x1
1 2 1 4
0 5 3 x2 1.
x
3
----(3)
Equation (3) has the form Ax b. Such an equation
is called a matrix equation.
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MATRIX EQUATION Ax b
Theorem 3: If A is an m n matrix, with columns
m
a1, …, an, and if b is in
, then the matrix equation
Ax b
has the same solution set as the vector equation
x1a1 x2a 2 ... xn an b ,
which, in turn, has the same solution set as the system
of linear equations whose augmented matrix is
a n b.
a1 a 2
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EXISTENCE OF SOLUTIONS
The equation Ax b has a solution if and only if b
is a linear combination of the columns of A.
Theorem 4: Let A be an m n matrix. Then the
following statements are logically equivalent. That
is, for a particular A, either they are all true
statements or they are all false.
m
a. For each b in
, the equation Ax b has a
solution.
b. Each b in m is a linear combination of the
columns of A.
m
c. The columns of A span .
d. A has a pivot position in every row.
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PROOF OF THEOREM 4
Statements (a), (b), and (c) are logically equivalent.
So, it suffices to show (for an arbitrary matrix A) that
(a) and (d) are either both true or false.
Let U be an echelon form of A.
m
Given b in , we can row reduce the augmented
matrix A b to an augmented matrix U d for
m
some d in :
A
b ...
U
d
If statement (d) is true, then each row of U contains a
pivot position, and there can be no pivot in the
augmented column.
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PROOF OF THEOREM 4
So Ax b has a solution for any b, and (a) is true.
If (d) is false, then the last row of U is all zeros.
Let d be any vector with a 1 in its last entry.
Then U d represents an inconsistent system.
Since row operations are reversible, U d can be
transformed into the form A b.
The new system Ax b is also inconsistent, and (a)
is false.
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COMPUTATION OF Ax
3 4
2
5 3
Example 2: Compute Ax, where A 1
x1
6 2 8
and x x2 .
x3
Solution: From the definition,
3 4 x1
2
2
3
4
1 5 3 x x 1 x 5 x 3
2 1 2 3
6 2 8 x3
6
2
8
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COMPUTATION OF Ax
2 x1 3 x2 4 x3
x1 5 x2 3 x3 ---(1)
6 x1 2 x2 8 x3
2 x1 3 x2 4 x3
x1 5 x2 3 x3
6 x1 2 x2 8 x3 .
The first entry in the product Ax is a sum of products
(a dot product), using the first row of A and the
entries in x.
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COMPUTATION OF Ax
2 3 4 x1 2 x1 3 x2 4 x3
x
.
That is,
2
x3
Similarly, the second entry in Ax can be calculated by
multiplying the entries in the second row of A by the
corresponding entries in x and then summing the
resulting products.
x1
1 5 3 x x 5 x 3 x
2
3
2 1
x3
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ROW-VECTOR RULE FOR COMPUTING Ax
Likewise, the third entry in Ax can be calculated from
the third row of A and the entries in x.
If the product Ax is defined, then the ith entry in Ax is
the sum of the products of corresponding entries from
row i of A and from the vertex x.
The matrix with 1s on the diagonal and 0s elsewhere
is called an identity matrix and is denoted by I.
1 0 0
For example, 0 1 0 is an identity matrix.
0 0 1
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PROPERTIES OF THE MATRIX-VECTOR
PRODUCT Ax
Theorem 5: If A is an m n matrix, u and v are
n
vectors in
, and c is a scalar, then
a. A(u v) Au Av;
b. A(cu) c( Au).
Proof: For simplicity, take n 3 , A a1 a 2 a 3 ,
3
and u, v in .
For i 1,2,3, let ui and vi be the ith entries in u and
v, respectively.
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PROPERTIES OF THE MATRIX-VECTOR
PRODUCT Ax
To prove statement (a), compute A(u v) as a linear
combination of the columns of A using the entries in
u v as weights.
A(u v) a1 a 2
u1 v1
a 3 u2 v2
u3 v3
Entries in
(u1 v1 )a1 (u2 v2 )a 2 (u3 v3 )a 3
uv
Columns of A
(u1a1 u2a 2 u3a 3 ) (v1a1 v2a 2 v3a 3 )
Au Av
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PROPERTIES OF THE MATRIX-VECTOR
PRODUCT Ax
To prove statement (b), compute A(cu) as a linear
combination of the columns of A using the entries in cu as
weights.
cu1
A(cu) a1 a 2 a 3 cu2 (cu1 )a1 (cu2 )a 2 (cu3 )a 3
cu3
c(u1a1 ) c(u2a 2 ) c(u3a 3 )
c(u1a1 u2a 2 u3a 3 )
c( Au)
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