```Website owner:  James Miller
```

[ Home ] [ Up ] [ Info ] [ Mail ]

Solution of a consistent system of linear equations

Solution of a consistent system of linear equations. The following theorem gives the complete solution of a consistent system of linear equations Ax = b. It deals not only with the case in which the rank of the matrix A is equal to the number of unknowns, in which case there is one single solution. It also deals with the case of underdetermined systems where the rank of A is less than the number of unknowns. The complete solution to the system in this case consists of any particular solution of the system Ax = b plus all solutions of the system Ax = 0 (a linear manifold of solutions).

Theorem. Let Ax = b be a consistent system having n unknowns, and let the rank of A be r.

Case 1. r = n. There is a single solution vector x which can be found by one of the usual methods.

Case 2. r < n. Let xp be a particular solution to the system Ax = b i.e. vector xp can be any vector that satisfies the system. Then the complete solution of the system Ax = b can be written as

X = xp + c1u1 + c2u2 + .... + cn-run-r

where c1, c2, .... ,cn-r are arbitrary constants and the vectors u1, u2, .... ,un-r are any set of linearly independent vectors that span the solution space of the system Ax = 0. There are n-r such vectors. In other words, one can find n-r linearly independent vectors u1, u2, .... ,un-r which satisfy the set of homogeneous equations Ax = 0. The vector xp plus any linear combination of these vectors u1, u2, .... ,un-r is a solution of the given equation. There are no other solutions. If b = 0, the vector xp can be taken as xp = 0.

Example. Solve the system:

We divide the first equation by 4 and use it to eliminate x1 from the remaining equations:

Now we rearrange to put the largest element in the proper position:

We divide the second equation by 4.5 and use it to eliminate x3 from the other two equations:

We see that the rank of A is 2 and that the system has two equations.

Because there are four unknowns and the rank of A is 2, we know from the theorem that the complete solution is made up of a particular solution and any linear combination of two linearly independent solution vectors.

We can find the complete solution as follows:

Add .25 times row 2 to row 1, obtaining:

We can find the particular solution by choosing two arbitrary values for x2 and x4. Let us choose x2 = x4 = 0. Substituting these values into the system we get

x1 = 13/9

x3 = -2/9

Thus the particular solution is:

To find two linearly independent solution vectors, we take the homogeneous equation

and choose arbitrary values for x2 and x4.

For the first vector let us take x2 = 0, x4 = 3. On substituting into the system we get

+ 1/3 = 0

- 5/3 = 0

which has the solution

x1 = -1/3, x3 = 5/3 .

Thus the first vector is:

For the second vector let us pick the values x2 = 3, x4 = 0. On substituting into the system we get:

+ 1 = 0

- 2 = 0

which has the solution

= -1, = 2 .

The second vector is then

and the general solution is:

where c1 and c2 are arbitrary constants.