Exceptionally, I have not included a video or homework for this Lesson. You can practise the method by using it to solve examples from next Lesson (on Gaussian elimination).

Instead, you will find here an article in which I present how the rank of a matrix can be used to solve systems of linear equations. In this article I assume that you already know how to compute the rank of a matrix and how to solve systems of equations using Cramer’s rule.

Topic: Using ranks of matrices for a system of linear equations

Using ranks of matrices for a system of linear equations

The idea of using matrices to study a system of linear equations is really very simple. Consider an arbitrary system of linear equations (this is very important- the number of unknowns does not have to be same as the number of equations):

Suppose our system has m equations and n unknowns. The coefficient matrix is the matrix formed from the coefficients of the unknowns. The rank of the coefficient matrix is simply the rank the following matrix:

The augmented matrix is obtained by adding to the coefficient matrix one extra column containing constant terms (the numbers on the right-hand side of the equations). The rank of the augmented matrix is the rank of this larger matrix.

The key fact: A system of linear equations has at least one solution if and only if where A is the coefficient matrix and (A|b) is the augmented matrix. . From this statement we obtain the following conclusion:

1. If the rank of the coefficient matrix, the rank of augmented matrix and the number of unknowns in the system are all equal (), then the system has exactly one solution.
2. If the rank of the coefficient matrix is equal to the rank of the augmented matrix, but this common rank is smaller than the number of unknowns () then the system has infinitely many solutions.
3. If the rank of the coefficient matrix is different from the rank of the augmented matrix (), then the system has no solutions.

How to use rank of matrices to solve a system of linear equations

We already know the relationship between ranks of matrices and solution of system of linear equations. The question is how to apply it in practice. The most straightforward, although somewhat time consuming, method is to compute both ranks separately-the rank of the coefficient matrix and the rank of the augmented matrix ).

Then we interpret the result, “reduce” the system to be able to apply Cramer’s rule (possibly omitting some equations and replacing certain unknowns with parameters), and solve using Cramer’s rule. I will demonstrate this method further in the article.

It is also possible to compute both ranks simultaneously using one matrix, to eliminate rows or columns at the same time, or to use Gaussian elimination instead. Sometimes it seems to me that there are as many methods as there are lecturers. Of course, all of them are correct, if they lead to the desired result — solving the system.

Example

We have to solve the system of linear equations shown above. First, of course, we check whether it can be solved using Cramer’s rule— that is, whether the number of equations equals the number of unknowns and whether the main determinant is non-zero. It clearly cannot be solved using Cramer’s rule, because there are three equations and four unknowns. Therefore, we do not solve it using Cramer’s rule; instead, we proceed to the ranks of matrices and use the result involving ranks. First, we compute the rank of the coefficient matrix:

We compute, compute, compute, just as one normally computes the rank of a matrix (I invite you, for example, to my Course – it is simple), and we obtain:

Now, we compute the rank of the augmented matrix:

Again, we compute, compute, compute, and obtain:

Thus, we obtain the situation: . Rank of the coefficient matrix is equal to the rank of the augmented matrix, and both are 3 (this is important). So, the system has solutions, and we continue the calculations. We write the coefficient matrix once again:

And now we choose from it any submatrix of the order, . In our case the rank of both the coefficient matrix and the augmented matrix turned out to be 3, so we choose any submatrix of order 3×3 and compute its determinant – but note – it must be a non-zero determinant (we have to calculate it and check this separately). We put the chosen matrix in a box.

We now form a system of equations consisting only of those equations whose rows appear in our matrix (the remaining equations are not written at all), and only of those unknowns whose columns appear in our matrix (the remaining unknowns are replaced by parameters). In our example we obtain a system consisting of the first, second and third equation (because the first, second and third row are contained in the matrix):

As it happens, these will be all the equations. As for the unknowns, we look at the columns that enter the chosen matrix:

These are the first, second and third unknown: . . The fourth unknown, namely did not “get in”. We replace it by a parameter, say , where may take any value, that is, it is arbitrary . Parameters can be denoted by various other letters, for example “t”, or we may not change the symbols at all and simply start treating certain variables as parameters without altering their notation We now form a new system of equations:

In this system we treat the parameters as numbers, so we move the corresponding terms to the right-hand side:

This is a system with same number of unknowns as equations, and we solve it using Cramer’s rule. How do we construct the determinants for the successive variables? Simply treat, for example, an expression such as: as a single number. For example:

And its values is: . We obtain the solution:

for .

Articles and blog posts related to this Lesson

• “Matrix rank in linear system of equations with a parameter”
• “System of Equations with a Parameter to Determine“
• “Homogeneous systems of linear equations (number of solutions using matrix rank)“
• “Gaussian vs Cramer’s vs Kronecker-Capelli Methods – Matrices in Solving Systems of Linear Equations“
• “The Biggest Problem with Matrix Problems…“