Systems of Linear Equations in Vector Form Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Systems of Linear Equations

in Vector Form

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Any system of linear equations can be put into matrix form:

Or vector form:

bxA



The vectors in the vector form of the equation are the columns of the matrix A in the matrix format.

Here are a few examples:

0x2x2x

0xx3x2

0xxx

321

321

321

3xx

1xx

1xxx

31

21

321

4xx4x

2xxx2x

1xx2x

532

5432

421

baxaxax nn2211

This is a HOMOGENEOUS system because the right side is all 0.

This is a NON-HOMOGENEOUS system because the right side is not all 0.

This is a NON-HOMOGENEOUS system because the right side is not all 0. Here we have fewer equations than unknowns.




0x2x2x

0xx3x2

0xxx

321

321

321

baxaxax nn2211

0

0

0

b;

2

1

1

a;

2

3

1

a;

1

2

1

a 321

Here are the vectors for this system.

0221

0132

0111




0x2x2x

0xx3x2

0xxx

321

321

321

baxaxax nn2211

0

0

0

b;

2

1

1

a;

2

3

1

a;

1

2

1

a 321


0221

0132

0111Row reduction yields:

The row of zeroes indicates a free variable, and an infinite # of solutions

0000

0310

0401




0x2x2x

0xx3x2

0xxx

321

321

321

baxaxax nn2211

0

0

0

b;

2

1

1

a;

2

3

1

a;

1

2

1

a 321


0221

0132


0000

0310

0401

33

32

31

xx

x3x

x4x

The row of zeroes indicates a free variable, and an infinite # of solutions

Here is the corresponding solution.

t

1

3

4

x

x

x

3

2

1

Rename X3, call it ‘t’, and we get a vector solution (this is a 1-dimensional subset of ℝ3):

This plane will also be called the Column Space of matrix A.

It is also the Span of the set



0x2x2x

0xx3x2

0xxx

321

321

321

baxaxax nn2211

0

0

0

b;

2

1

1

a;

2

3

1

a;

1

2

1

a 321


Because we got a free variable in our row reduction process, we conclude that vectors a1, a2 and a3 are linearly dependent.

Furthermore, since we got 2 pivots in our reduced matrix, we can say that these 3 vectors span a 2-dimensional subset of ℝ3 (a plane, pictured below).

a1

a2

a3

321 a,a,a



0x2x2x

0xx3x2

0xxx

321

321

321

baxaxax nn2211

0

0

0

b;

2

1

1

a;

2

3

1

a;

1

2

1

a 321


Because we got a free variable in our row reduction process, we have infinitely many solutions to the system. The set of all solutions form a 1-dimensional subspace of ℝ3. Since this system is homogeneous, we call this solution set the Null Space of matrix A.

The solution was written as a vector. The Null Space consists of all multiples of this vector. Geometrically, this space is a line in ℝ3, pictured below.

5

0

5

5 0 5 2

1

0

1

2

1

3

4




baxaxax nn2211

3

1

1

b;

1

0

1

a;

0

1

1

a;

1

1

1

a 321


3101

1011

1111

3xx

1xx

1xxx

31

21

321




baxaxax nn2211

3

1

1

b;

1

0

1

a;

0

1

1

a;

1

1

1

a 321


3101

1011


2100

2010

1001

3xx

1xx

1xxx

31

21

321




baxaxax nn2211

3

1

1

b;

1

0

1

a;

0

1

1

a;

1

1

1

a 321


3101

1011


2100

2010

1001

2x

2x

1x

3

2

1

Here is the unique solution:

3xx

1xx

1xxx

31

21

321

This solution tells us the specific linear combination of a1 a2 and a3 that adds up to the right side vector b.

3

1

1

1

0

1

)2(

0

1

1

)2(

1

1

1

)1(



4xx4x

2xxx2x

1xx2x

532

5432

421

baxaxax nn2211



4

2

1

10410

11210

01021

4

2

1

b

1

1

0

a;

0

1

1

a;

4

2

0

a;

1

1

2

a;

0

0

1

a 54321



4xx4x

2xxx2x

1xx2x

532

5432

421

baxaxax nn2211



4

2

1

10410

11210


1

0

1

0100

1010

2001

6132

31

4

2

1

b

1

1

0

a;

0

1

1

a;

4

2

0

a;

1

1

2

a;

0

0

1

a 54321



4xx4x

2xxx2x

1xx2x

532

5432

421

baxaxax nn2211



4

2

1

10410

11210


1

0

1

0100

1010

2001

6132

31

Here is the corresponding solution.

0x1x0x

0x0x1x

1x0xx

0x1xx

1x2xx

545

544

5461

3

5432

2

5431

1

4

2

1

b

1

1

0

a;

0

1

1

a;

4

2

0

a;

1

1

2

a;

0

0

1

a 54321



4xx4x

2xxx2x

1xx2x

532

5432

421

baxaxax nn2211


4

2

1

b

1

1

0

a;

0

1

1

a;

4

2

0

a;

1

1

2

a;

0

0

1

a 54321


4

2

1

10410

11210


1

0

1

0100

1010

2001

6132

31

Here is the corresponding solution. There are 2 free variables, so we get a 2-dimensional subset of ℝ5.

0

0

1

0

1

t

1

0

0

1

2

s

0

1

x613231

0x1x0x

0x0x1x

1x0xx

0x1xx

1x2xx

545

544

5461

3

5432

2

5431

1

Documents

Systems of Linear Equations in Vector Form Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB