Slide 1.4

1 Linear Equations
in Linear Algebra
1.4
THE MATRIX EQUATION Ax  b
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.
Slide 1.4- 2
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
v2
 3
v3   5   Ax .
 
 7 
Slide 1.4- 3
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
----(2)
Slide 1.4- 4
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.
Slide 1.4- 5
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
Slide 1.4- 6
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.
Slide 1.4- 7
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.
Slide 1.4- 8
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.





Slide 1.4- 9
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
Slide 1.4- 10
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.
Slide 1.4- 11
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  

Slide 1.4- 12
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
Slide 1.4- 13
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.
Slide 1.4- 14
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
uv
Columns of A
 (u1a1  u2a 2  u3a 3 )  (v1a1  v2a 2  v3a 3 )
 Au  Av
Slide 1.4- 15
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)