The rank of the matrix in the Kronecker-Capelli theorem

Rank of matrices Lecture 3

Topic: Kronecker-Capelli theorem

Summary

In this article, I will show how the order of the matrix is used in solving systems of linear equations using the Kronecker-Capelli method (more correctly: a method using the Kronecker-Capelli theorem). In the article, I assume that you already know how to calculate the rank of matrices and systems of equations with Cramer’s formulas.

Kronecker-Capelli theorem

The Kronecker-Capelli theorem using the rank of a matrix is really very simple. Having an arbitrary (this is super-important, meaning that the variables do not have to be as many as the equations) system of equations:

Our system – pay attention – has m equations and n variables. The rank of the principal matrix is the rank of the matrix formed from the coefficients at the variabels, that is:

Of course, it does not have to be a square matrix. The rank of a completed matrix is the rank of a matrix formed from the coefficients at the variables with an added column of free expressions (on the right sides of the equations):

The Kronecker-Capelli theorem states that a system has solutions if and only if the rank of the principal matrix equals the rank of the complemented matrix:

rz(A)=rz(U)

The following conclusions follow from the statement:

If the rank of the principal matrix, the rank of the complementary matrix and the number of variables in the system are equal ( $rz(A)=rz(U)=n$ ) then the system of equations has exactly one solution.
If the rank of the principal matrix is the same as the rank of the complementary matrix, but is smaller than the number of variables ( $rz(A)=rz(U)<n$ ) then the system of equations has infinitely many solutions.
If the rank of the principal matrix is different from the rank of the complementary matrix ( $r z open parentheses A close parentheses not equal to r z open parentheses U close parentheses$ ), then the system of equations has no solutions.

Application of the Kronecker-Capelli theorem (method of solving a system of equations)

The theorem is known. The question remains how to apply them in practice.

The most “clean” (but, unfortunately, also time-consuming) method is to simply count both ranks (the rank of the main matrix rank ( $A$ ) and the rank of the supplemented matrix rank ( $U$ ) completely separately, finally interpret the result, “truncate” the system to a Cramer system (by possibly plotting some equations and replacing some variables with parameters) and solve the resulting Cramer system. I will show you this method further on in the article.

You can also count both ranks simultaneously on a single matrix, you can zero rows or columns at the same time, you can count properly using the Gaussian method…. Sometimes it seems to me that as many professors as many methods. Of course, they are all good, as long as they lead to the goal of solving the system.

Example

We have the above system of equations to solve. First, of course, we check that it is not a Cramer system, i.e. that it has as many equations as variables and that the system’s principal determinant is different from zero. Of course, this is not a Cramer system, because we have 3 equations and 4 variables in it. So we don’t solve the system with Cramer’s formulas at this point, but move on to matrix orders and the Kronecker-Capelli theorem.

First, we count the rank of the main matrix, that is:

We count, count, count, the way one counts ranks of matrices (I invite you to take a look at my Course for example – it’s really easy) and we have the result:

Now we count the rank of the complemented matrix:

We count, we count, we count and we have the result:

Thus, we have a situation:

rz(A)=rz(U)=3

The rank of the main matrix is equal to the rank of the complemented matrix and they are equal to 3 (this is important). That is, the system will have a solution and we continue counting. We write the main matrix again:

And now we choose from it any determinant of degree $rz(A)=rz(U)$ . In our case, the rank of the principal and complementary matrix came out equal to 3. That is, we choose any determinant of the 3rd degree – but note – it must be a determinant different from zero (you need to count and check on the side). We take the selected determinant in a frame:

Now we create a system of equations only from the equations whose rows are in our determinant (we do not write the other equations at all) and only from the variables whose columns are in our determinant (we replace the other variables with parameters).

In our example, we will create a system of equations consisting of equation one, two and three (because the first, second and third lines were in the determinant):

As it happens, they will be all equations.

As for the variables, we look at the columns that got to the chosen determinant:

This is the first, second and third variables: $x_1, x_2, x_3$ . “Did not catch” the fourth variable, which is $x_4$ . We replace it with the parameter: $x_4={\alpha} _1$ where ${\alpha}_1$ takes any value, i.e. ${\alpha}_1 {\in}{\R}$ . You can label parameters with various other letters, such as “t”, or you can not label them with letters at all, but just start treating them as parameters without changing the designations.

We create a new system of equations:

In it, we treat the parameters as numbers, that is, we flip to the right side:

This is a Cramer system and we solve it with Cramer’s formulas. How to create determinants for the following variables? Simply treat for example: $1-7{\alpha} _1$ – as one number. For example:

and its value: $-8+{\alpha}_1$ .