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Technique for solving underdetermined systems of linear equations

We wish to demonstrate the technique for finding the full solution to the linear system Ax = b of m equations in n unknowns where the rank r of matrix A is less than or equal to the number of unknowns n. When r is less than n the system is said to be underdetermined. The technique is based on the following theorem:

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. Let us suppose that the system Ax = b consists of the single equation

5x + 3y + 9z = 13

This equation corresponds to a plane in three-dimensional space. Envision this plane in space and label it "K". This plane represents the solution set of the system. Pass a plane through the origin parallel to plane K and label it L. Plane L represents the solution set to the homogeneous system AX = 0 i.e. the equation

5x + 3y + 9z = 0

In this case the rank r of the coefficient matrix A is 1 and the number n of unknowns is 3. We need a particular solution for the system Ax = b and any two linearly independent vectors to serve as a basis for the solution space of Ax = 0. Plane L represents the solution space of the system Ax = 0. So what we need is the coordinates of any arbitrarily chosen point on plane K to serve as our particular solution and the coordinates of any two points on plane L to give us two linearly independent vectors to serve as a basis for the solution space of Ax = 0 (making sure the vectors are not collinear).

Computing the particular solution. We can compute the coordinates of some point on plane K by arbitrarily picking values for a couple of the unknowns and solving for the other unknown. Let us choose x = 1 and y = 1. Then and z = 5/9. So a particular solution is

Computing basis vectors for the solution space of Ax = 0. Let us select a couple of points on plane L to serve as basis vectors for the solution space of Ax = 0.

Point 1. Choose x = 1, y = 1. Then 5∙1 + 3∙1 + 9z = 0 and z = -8/9.

Coordinates of Point 1: (1, 1, -8/9)

Point 2. Choose x = -2, y = -3. Then 5(-2) + 3(-3) + 9z = 0 and z = 19/9.

Coordinates of Point 2: (-2, -3, 19/9)

Full solution of the system:

where c1 and c2 are arbitrary constants.

Because of the very simple example that we used there was no need to reduce matrix A to row canonical form. In most problems the first step would be to reduce A to row canonical form by using elementary row operations. Then, after doing that, the procedure would follow that which we have outlined in this simple example. The particular solution consists of any point that satisfies the system Ax = b (or its equivalent reduced system Cx = d). The basis vectors for the solution space of Ax = 0 are found by computing n-r points that satisfy Ax = 0 (or its equivalent reduced system Cx = 0).