4Matrices and linear equations
IA Vectors and Matrices
4.6 General solution of Ax = d
Finally consider the general equation
Ax
=
d
, where
A
is an
n × n
matrix and
x, d are n × 1 column vectors. We can separate into two main cases.
(i) det
(
A
)
6
= 0. So
A
−1
exists and
n
(
A
) = 0,
r
(
A
) =
n
. Then for any
d ∈ R
n
,
a unique solution must exists and it is x = A
−1
d.
(ii) det
(
A
) = 0. Then
A
−1
does not exist, and
n
(
A
)
>
0,
r
(
A
)
< n
. So the
image of A is not the whole of R
n
.
(a) If d 6∈ im A, then there is no solution (by definition of the image)
(b)
If
d ∈ im A
, then by definition there exists at least one
x
such that
Ax
=
d
. The general solution of
Ax
=
d
can be written as
x
=
x
0
+
y
,
where
x
0
is a particular solution (i.e.
Ax
0
=
d
), and
y
is any vector
in ker A (i.e. Ay = 0). (cf. Isomorphism theorem)
If
n
(
A
) = 0, then
y = 0
only, and then the solution is unique (i.e.
case (i)). If
n
(
A
)
>
0 , then
{u
i
}, i
= 1
, ··· , n
(
A
) is a basis of the
kernel. Hence
y =
n(A)
X
j=1
µ
j
u
j
,
so
x = x
0
+
n(A)
X
j=1
µ
j
u
j
for any µ
j
, i.e. there are infinitely many solutions.
Example.
1 1
a 1
x
1
x
2
=
1
b
We have det A = 1 − a. If a 6= 1, then A
−1
exists and
A
−1
=
1
1 − a
=
1
1 − a
1 −1
−a 1
.
Then
x =
1
1 − a
1 − b
−a + b
.
If a = 1, then
Ax =
x
1
+ x
2
x
1
+ x
2
= (x
1
+ x
2
)
1
1
.
So
im A
=
span
1
1
and
ker A
=
span
1
−1
. If
b 6
= 1, then
1
b
6∈ im A
and there is no solution. If b = 1, then
1
b
∈ im A.
We find a particular solution of
1
0
. So The general solution is
x =
1
0
+ λ
1
−1
.
Example. Find the general solution of
a a b
b a a
a b a
x
y
z
=
1
c
1
We have
det A
= (
a − b
)
2
(2
a
+
b
). If
a 6
=
b
and
b 6
=
−
2
a
, then the inverse exists
and there is a unique solution for any c. Otherwise, the possible cases are
(i) a
=
b, b 6
=
−
2
a
. So
a 6
= 0. The kernel is the plane
x
+
y
+
z
= 0 which is
span
−1
1
0
,
−1
0
1
We extend this basis to R
3
by adding
1
0
0
.
So the image is the span of
a
a
a
=
1
1
1
. Hence if
c 6
= 1, then
1
c
1
is not
in the image and there is no solution. If
c
= 1, then a particular solution
is
1
a
0
0
and the general solution is
x =
1
a
0
0
+ λ
−1
1
0
+ µ
−1
0
1
(ii) If a 6= b and b = −2a, then a 6= 0. The kernel satisfies
x + y − 2z = 0
−2x + y + z = 0
x − 2y + z = 0
This can be solved to give
x
=
y
=
z
, and the kernel is
span
1
1
1
. We
add
1
0
0
and
0
0
1
to form a basis of
R
3
. So the image is the span of
1
−2
1
,
−2
1
1
.
If
1
c
1
is in the image, then
1
c
1
= λ
1
−2
1
+ µ
−2
1
1
.
Then the only solution is
µ
= 0
, λ
= 1
, c
=
−
2. Thus there is no solution if
c 6
=
−
2, and when
c
=
−
2, pick a particular solution
1
a
0
0
and the general
solution is
x =
1
a
0
0
+ λ
1
1
1
(iii)
If
a
=
b
and
b
=
−
2
a
, then
a
=
b
= 0 and
ker A
=
R
3
. So there is no
solution for any c.