1.4 The Matrix Equation Ax = b
We can write linear combinations as
multiplication of a matrix with a vector.
Definition: If A is an m x n matrix with
columns a1, a2 ,, an and if x is in Rn,
then the product of A and x denoted by
Ax is a linear combination or the
columns of A using the corresponding
entries of x as weights.
That is,
 x1 
Ax  a1 a2  an   
 xn 
 x1a1  x2a2    xn an
For example,
1  4
1
 4
3 2  7   7 3   6  2

  6
 
 


0 5
0
 5
 7   24   31 
 21   - 12    9 
 0  - 30  30
Example: Consider:
 2 3 4 9
  3 1 0  2


We can write the system of equations
corresponding to the augmented matrix:
2 x1  3x2  4 x3  9
 3x1  x2
 2
Then express the system of equations in
vector form
 2
 3
 4  9 
x1    x2    x3     
  3
 1
 0    2
Then express this vector equation in the
form Ax = b where b is a 2 x 1 vector.
 x1 
 2 3 4    9 
  3 1 0   x2    2

x   
 3
►We now have three equivalent ways of
viewing a linear system:
1. as a system of linear equations
2. as a vector equation
3. as a matrix equation
Theorem 3
If A is an m x n matrix with columns
a1, a2 ,, an and if b is in Rm, then the
matrix equation
Ax = b
has the same solution set as the vector
equation is
x1v1  x2 v 2    xn v n  b
which in turn has the same solution set
as the system of linear equations whose
augmented matrix is
a1 a2 a3 b .
Useful fact: The equation Ax = b has a
solution if an only if b is a linear
combination of the columns of A.
Example: Let
 b1 
 1 4 5
 
A    3  11 14 b  b2 
b3 
 2 8 10 ,
Is Ax = b consistent for all b?
Create the augmented matrix
corresponding to Ax = b.
 1 4 5 b1 
  3  11  14 b 
2

 2 8 10 b3 
R2+3R1
R3 + (-2R1)
 1
 0

 0
4
5
1 1
0 0


b2  3b1 
b3  2b1 
b1
Is this consistent for all b?
There are values of b1 ,b3 for which
b3  2b1  0 . So, Ax = b in inconsistent
for some b.
Note:
b3  2b1  0 is a plane in R3.
If any b in R3 can be expressed as linear
combination of the columns of A, then
we say the columns of A span R3.
Definition: We say that the columns of
a1 a2  a p  span Rm if every vector b in
Rm is a linear combination of
a1 , a 2 ,  , a p ,
and Span{ a1 , a 2 ,  , a p }= Rm.
Theorem 1.4
Let A be an m x n matrix. Then the
following statements are logically
equivalent. (That is, for a particular A,
either they are all true or they are all
false.)
a) For each b in Rm, Ax = b has a
solution.
b) Each b in Rm is a linear
combination of the columns of A.
c) The columns of A span Rm.
d) A has a pivot position in every row.
Proof: c ↔ b: The columns of A span
Rm if and only if each b in Rm is a linear
combination of the columns of A, by the
definition of span.
b ↔ a: Each b in Rm is a linear
combination of the columns of A if and
only if the equation Ax = b has a solution
for each b in Rm, by the Useful Fact.
We have c ↔ b and b ↔ a so c ↔ a
All we need to show is d ↔ a, b, or c.
d ↔ a is easiest.
First d → a:
A has a pivot position in every row.
Let U be a row echelon form of A.
Given b in Rm, we can write the
augmented matrix [A b] whose columns
are the columns of A and the vector b.
We can row reduce this matrix to [U d]
for some d in Rm.
[U d] is now an augmented matrix in row
echelon form.
A has a pivot position in every row, so U
has a pivot in every row. Thus, there is
no pivot in d (by Homework problem
1.2.25). Thus, by Th 1.2, the system
represented by [A b] is consistent, and
the equation Ax = b has a solution so (a)
is true.
a → d:
it is easier to prove the contrapositive:
If (d) is not true, then (a) is not true: that
is, we need to prove that if there is a row
in A that has no pivot position, there
exists a vector b in Rm, for which the
equation Ax = b has no solution. We
need to find that b.
Say (d) is false. Then there is a row in A
that does not have a pivot position, so
the last row in U is all zeros since U is a
row echelon form of A.
Let d be any vector with a 1 in its last
entry. Then [U d] represents an
inconsistent system since the far right
column has a pivot.
Recall: row operations are reversible, so
[U d] can be transformed into [A b] for
some b. The new system represented
by [A b] or Ax = b is inconsistent, so (a)
is false.
Since not d → not a, a → d, and we are
done.
QED
Example: Let
 b1 
1 2 




b  b2 
A  3 4 
b3 
5 6 and
Is Ax = b consistent for all possible b?
Th. 1.4 says the equation is consistent if
A has a pivot in each row. But, there are
three rows in A and only two columns, so
A can have at most two pivots.Thus, the
equation is inconsistent for all possible b.
Example: Do the columns of
1 2 3 


A   2 4 6
3
span
R
?
0 3 9
Th. 1.4 says A has a pivot position in
every row if and only if the columns of A
span R3.
Row 2 has no pivot since it is a multiple
of Row 1.
Or
1 0 - 3


rref  A  0 1 3 
0 0 0 
And Row 3 has no pivot.
Since there are only two rows with
pivots, the columns of A do not spen R3.
►There is another rule for computing
Ax. Read Examples 4 and 5 on p44-45
to see it.
Theorem 1.5: If A is an m x n matrix and
u and v are vectors in Rn, and c is a
scalar, then
a. A(u + v) = Au + Av;
b. A(cu) = c(Au)
Proof of a:
Au  v 
u1  v1 
u  v 
2
2

a1 a2  an 
  


un  vn 
u1  v1 a1  u2  v2 a2    un  vn a n
u1a1  v1a1    un an  vn an
u1a1    unan   v1a1    vnan 
Au  Av