I will now explain** how to know if a system of equations has a solution** and we will also look at the types of equation systems that exist, since both concepts are closely related. All with step-by-step exercises.

When you are going to solve a system of linear equations, the first question you should ask yourself is: does it have a solution?

It may be that you have a single solution, that you have no solution or that you have infinite solutions, depending on the type of system it is.

If it does, you can start solving it any way you want, but if it doesn’t have a solution, you won’t have to go on and you’ll have saved yourself a lot of time.

I will explain it to you at the same time as we see it with an example.

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## Matrix of coefficients and extended matrix

In order to determine the type of a system of linear equations, or what it is, if a system of equations has a solution, we must be clear about the matrix of coefficients and the extended matrix.

Let’s see it in the next system:

First we must distinguish between the matrix of coefficients, which we will call matrix A and the extended matrix, which we will call matrix A*.

### What is the matrix of coefficients

The matrix of the coefficients, the matrix A, is the one formed by the coefficients that are multiplying the unknowns.

For the example system, the matrix A of the coefficients is:

Notice that the unknowns are in order: first the x’s, then the y’s and then the z’s.

Besides, if there are any unknowns missing from an equation, it will have a coefficient of zero.

### What is the expanded matrix

The expanded matrix, matrix A*, is formed by adding a column to the right of the matrix of coefficients, the elements of which are formed by the terms in the second member of each equation

The expanded matrix of our system is:

## Types of linear equation systems

How do you know if a system of linear equations has a solution or not?

We will know if a system of linear equations has a solution or not, comparing the ranges of the matrix of the coefficients and of the extended matrix with the number of unknowns, according to the Rouché-Fröbenius theorem, which says so:

### Rouché-Fröbenius’s Theorem

A system of linear m equations with m unknowns is compatible (has a solution) if, and only if, the range of the coefficient matrix coincides with the range of the extended matrix.

Depending on the number of solutions, the system will be of one type or another.

Let’s go see them.

### Determined compatible system

If the range of the coefficient matrix is equal to the range of the expanded matrix and is also equal to the number of unknowns in the system, then the system is determined compatible:

A system is compatible determined when it has a single solution. In this case, you can solve the system.

### System compatible undetermined

If the range of the coefficient matrix is equal to the range of the expanded matrix, but not equal to the number of unknowns in the system, then the system is indeterminate compatible:

A system is indeterminate compatible when it has infinite solutions, so you can no longer solve the system.

### Incompatible system

If the range of the coefficient matrix and the range of the extended matrix are not the same, the system is incompatible and has no solution:

## Solved exercises of systems of linear equations 3×3

Let’s see if the system of linear equations in the example has a solution.

We calculated the range of A. The largest submatrix is the A matrix itself:

Whose determinant is zero, so the range of A is going to be less than 3.

We choose the order 2 sub-matrix formed by columns 1 and 2 and rows 1 and 2. Its determinant is:

It is non-zero, so the range of matrix A is 2:

Now let’s calculate the range of the extended matrix.

The first sub-matrix of order 3 that can be chosen coincides with matrix A, which is formed by columns 1, 2 and 3, therefore we already know that the determinant of this sub-matrix is zero.

We are still testing the sub-matrix formed by columns 1, 2 and 4:

Its determinant is zero.

We tried the submatrix formed by columns 1, 3 and 4:

Its determinant is also zero.

And all we can do is try the submatrix formed by columns 2, 3 and 4:

All the determinants of the order 3 sub-matrixes contained in the extended matrix are zero, so the range of the extended matrix will be less than 3:

We choose a sub-matrix of order 2 consisting of columns 1 and 2 and rows 1 and 2:

Whose determinant is not zero, so the range of the extended matrix is 2

The range of the matrix of coefficients and of the extended matrix is the same and is equal to 2, but less than the number of unknowns, which is equal to 3, so the system is compatible indeterminate.

And therefore it has infinite solutions and cannot be solved.