Add the linear expressions
2 answers:
You might be interested in
Answer:
a) The system has a unique solution for and any value of
, and we say the system is consisted
b) The system has infinite solutions for and
c) The system has no solution for and
Step-by-step explanation:
Since we need to base the solutions of the system on one of the independent terms (), the determinant method is not suitable and therefore we use the Gauss elimination method.
The first step is to write our system in the augmented matrix form:
The we can use the transformation , obtaining:
.
Now we can start the analysis:
- If
then, the system has a unique solution for any value of
, meaning that the last row will transform back to the equation as:
from where we can see that only in the case of the value of
can not be determined.
- if
and
the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:
which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get
as a function of
o viceversa. Thus,
(
) is called a parameter since there are no constraints on what values they can take on.
if and
the system has no solution. Again by substituting in the equation resulting from the last row:
which is false for all values of
and since we have something that is not possible
the system has no solution