Possible, undetermined and impossible systems of linear equations

[diogomgf]

In the following system presented as an augmented matrix:



The book says the following:
1.If [MATH]a \neq 0[/MATH] and [MATH]b \neq 3[/MATH], the system is possible and determined.
2.If [MATH]a = 0[/MATH] and [MATH]b \in R[/MATH] the system is impossible.
3.If [MATH]a = 1[/MATH] and [MATH]b = 3[/MATH] the system is possible undetermined.
4.If [MATH]a \neq 0[/MATH] and [MATH]a \neq 1[/MATH] and [MATH]b = 3[/MATH] the system is impossible.


Point 1 is easy to understand, but points 2 trough 4 I don't fully understand.
Could some one help me out?

 

[Romsek]

I think the clearest way to address this is to plug in what values we are given and see where the Gaussian reduction takes us.
[MATH] a=0,~b \in \mathbb{R}\\ \begin{pmatrix} 1 &0 &0 &|&3\\ 1&b-3 &0 &|&3\\ 1 &0 &0 &| &4 \end{pmatrix}\\ \begin{pmatrix} 1 &0 &0 &|&3\\ 0&b-3 &0 &|&0\\ 0 &0 &0 &| &1 \end{pmatrix}\\ [/MATH]
and immediately you see the system is impossible. There is no linear combination of 0 that will result in 1.

3)
[MATH]a=1,~b=3\\ \begin{pmatrix} 1 &0 &1 &|&3\\ 1&0 &2 &|&4\\ 1 &0 &2 &| &4 \end{pmatrix}\\ \begin{pmatrix} 1 &0 &1 &|&3\\ 0&0 &1 &|&1\\ 0 &0 &0 &| &0 \end{pmatrix}\\ \begin{pmatrix} 1 &0 &0 &|&2\\ 0&0 &1 &|&1\\ 0 &0 &0 &| &0 \end{pmatrix}\\ x=2,~y\in \mathbb{R},~z=1\\ \text{The system is undetermined} [/MATH]
Do you think you can work out point 4?

 

[diogomgf]

I think the clearest way to address this is to plug in what values we are given and see where the Gaussian reduction takes us.

The problem is that nowhere in the exercise do they point out points 1 trough 4. That is the solution. The only thing asked is to "discuss the system of linear equations according to a and b"...
I worked them trough gauss-jordan elimination and posted my process, but I don't understand conclusions 2 trough 4 based on the augmented matrix obtained...

 

[pka]

In the following system presented as an augmented matrix:

View attachment 17448
The book says the following:
1.If [MATH]a \neq 0[/MATH] and [MATH]b \neq 3[/MATH], the system is possible and determined.
2.If [MATH]a = 0[/MATH] and [MATH]b \in R[/MATH] the system is impossible.
3.If [MATH]a = 1[/MATH] and [MATH]b = 3[/MATH] the system is possible undetermined.
4.If [MATH]a \neq 0[/MATH] and [MATH]a \neq 1[/MATH] and [MATH]b = 3[/MATH] the system is impossible.
Point 1 is easy to understand, but points 2 trough 4 I don't fully understand.

If you look at the determinate of the matrix of coefficients. It is \(ab-3a\).
With that in mind, look at each of the four given answers. Give yourself time to see the connection.

 

[diogomgf]

If you look at the determinate of the matrix of coefficients. It is \(ab-3a\).
With that in mind, look at each of the four given answers. Give yourself time to see the connection.

I don't know what the determinant of a matrix is...

 

[diogomgf]

I solved this exercise.
My main problem was that I was going directly to reduced row-echelon form.
If first I transform to row-echelon form and start analysing the systems based on the variables and then, if needed, reduce further all the steps in the O.P are covered.