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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 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:

      ole.gif     ole1.gif

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

             ole8.gif  + 1/3 = 0

            ole9.gif  - 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:

             ole11.gif  + 1 = 0

            ole12.gif  - 2 = 0

which has the solution

              ole13.gif  = -1, ole14.gif = 2 .

The second vector is then


and the general solution is:


where c1 and c2 are arbitrary constants.

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