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 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 be a particular solution to the system Ax = b i.e. vector can be any vector that satisfies the system. Then the complete solution of the system Ax = b can be written as

            X = +

where are arbitrary constants and the vectors 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 which satisfy the set of homogeneous equations Ax = 0. The vector plus any linear combination of these vectors is a solution of the given equation. There are no other solutions. If b = 0, the vector can be taken as = 0.

Example. Solve the system:

We divide the first equation by 4 and use it to eliminate 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 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 = 0. Substituting these values into the system we get

       = 13/9

       = -2/9

Thus the particular solution is:

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

and choose arbitrary values for and .

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

              + 1/3 = 0

             - 5/3 = 0

which has the solution

             = -1/3, = 5/3 .

Thus the first vector is:

For the second vector let us pick the values = 3, = 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 and are arbitrary constants.