﻿ ▷ Discussion of systems through determinants and ranges. Exercises

# Discussion of systems through determinants and ranges. Exercises resolved

Now I will explain how to discuss a system of three equations with three unknowns using determinants and ranges, with exercises solved step by step.

The discussion of linear equation systems is to analyze what type and how many solutions a system of linear equations has with three unknowns based on an unknown parameter.

## Discuss a system as a function of a parameter using determinants. Exercise solved

Next, we are going to solve an exercise on discussion of systems, using determinants. It is also possible to discuss the system using the Gauss method.

To understand this exercise you need to know how to identify the type of system by studying its ranges.

Discuss the following system of linear equations as a function of parameter a:

First, we get the matrix of the coefficients:

And the expanded matrix:

And now let’s calculate the determinant of the matrix of the coefficients, whose result will depend on parameter a:

To calculate it, taking advantage of the fact that I have a zero in column 2, I will add row 2 to row 1 to have another zero:

And it fits me:

I calculate the determinant from the attachments in the second column:

The result of the determinant of the matrix of coefficients has been left to me as a function of parameter a, as I mentioned earlier.

Now we are going to calculate the values of parameter “a” that make the determinant zero.

For them, we equal to zero the expression of the result of the determinant, which leaves us with a complete second degree equation, whose unknown is parameter a:

Whose solutions are..:

From now on, we start to discuss the system according to the values of parameter “a”.

### The determinant of A is other than zero

The values we have just calculated are those that make the determinant of A zero.

What if a≠ and a≠-3?

If a is not equal to 1, nor is it equal to -3, the determinant of A will be other than zero:

In this case, as the determinant of A is not zero, the range of the A matrix will be 3:

The range of the extended matrix will also be 3, as we will choose the square sub-matrix of order 3 that coincides with the matrix of the coefficients:

Then, as the range of the matrix coefficients and the range of the extended matrix are equal to 3 and also coincide with the number of unknowns, the system is compatible, determined and has a solution.

Therefore, for any value of parameter’a’ other than 1 and -3 the system will be compatible determined.

### The determinant of A is zero

The determinant of A will be zero when a=1 and when a=-3.

We are going to study each of the cases.

We begin by analyzing what type the system is when a=1:

We replace the parameter a with 1 both in the coefficient matrix and in the extended matrix and we are left with it:

We know that when a=1, the determinant of A is zero, therefore the range will be less than 3:

We choose the square sub-matrix of order 2 formed by columns 1 and 2 and rows 1 and 2 and calculate its determinant:

Its determinant is not zero, so the range of the matrix of coefficients is equal to 2:

Let’s now calculate the range of the extended matrix.

Since the extended matrix has a column of zeros, all the determinants of the sub-matrixes of order 3 of the extended matrix will be zero, so the range of the extended matrix will also be less than 3:

We continue testing with square submatrixes of order 2 and choose again the submatrix formed by columns 1 and 2 and rows 1 and 2 and whose determinant is not zero, so the range of the extended matrix is also 2:

Therefore, the range of the coefficient matrix is equal to the range of the expanded matrix, which is equal to 2, but is less than the number of unknowns, so the system is compatible indeterminate when a=1:

Let’s now see what type the system is when a=-3:

We replace the parameter a with -3 both in the coefficient matrix and in the extended matrix and we are left with it:

We know that when a=-3, the determinant of A is zero, therefore the range will be less than 3:

We choose the square sub-matrix of order 2 formed by columns 1 and 2 and rows 1 and 2 and calculate its determinant:

Its determinant is not zero, so the range of the matrix of coefficients is equal to 2:

We are now going to obtain the range of the extended matrix and for this purpose we calculate the determinant of the square submatrix of order 3 formed by columns 2, 3 and 4:

Which is non-zero, so the range of the extended matrix is equal to 3:

When a=-3, the range of the coefficient matrix and the range of the extended matrix are not the same, so the system is incompatible and has no solution: