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.