Linear system of equations: Echelon form/Solutions

[mathmari]

Hey!

I am looking at the following exercise but I think that I miss something.

The statement is the following:

We are given the following system of equations: \begin{align*}2a-2c+d-2e=&-2 \\ -2c-2d+2e=&\ \ \ \ \ 3 \\ d+2e=&-2\end{align*}

1) Is the system in echelon form? Justify.
2) Solve the linear system of equations over $\mathbb{R}$.
3) How many solutions has the system?
The system in matrix form is: \begin{equation*}\begin{pmatrix}\left.\begin{matrix}2 & -2 & 1 & -2 \\ 0 & -2 & -2 & 2 \\ 0 & 0 & 1 & 2\end{matrix}\right|\begin{matrix}-2 \\ 3\\ -2\end{matrix}\end{pmatrix}\end{equation*}
So the system is in echelon form, or not?

Isn't it trivial? Or do I miss something?

Does maybe the fact that the variable $b$ is missing important here?

:unsure:

 

[I like Serena]

Hey mathmari!

It looks like an introductory exercise for using echelon forms to solve a system of equations.
First it checks whether the word 'echelon' is understood, and then it asks to apply the corresponding method.
In other words, I don't think you're missing anything.

 

[HOI]

It doesn't matter at all what you call the letters representing the unknown numbers!

So, no, it doesn't matter that there is no "b". If the equations were
2a−2b+c−2d=-2
−2b−2c+2d= 3
c+2d=−2
The solution set would be exactly the same!

Personally, I would not bother writing the equations in matrix form.
The last equation, c+ 2d= -2 says that c= -2d- 2.
Then -2b- 2c+ d= 3 becomes -2b= 2c- d+ 3= -2d- 2- d+ 3= -3d+ 1 and then b= (3/2)d- 1/2.
And 2a- 2b+ c- 2d= -2 becomes 2a= 2b- c+ 2d- 2= 3d- 1+ 2d+ 2- 2d= 3d- 3 so
a= (3/2)d- 3/2.

Or, using the labels "a", "c", "d", and "e", as originally, and writing the solution in terms of parametera "t" (you have 4 unknowns with only three equations so the "degrees of freedom" is 4- 3= 1 and there is one parameter.)
a= (3/2)t- 3/2
c= (3/2)t- 3/2
d= - 2t- 2
e= t.

 

[Evgeny.Makarov]

The system is in row echelon form, but not in reduced row echelon form. In fact, some textbooks, (for example, Nicholson, W. Keith. Linear Algebra with Applications. 2019) require that the leading coefficients are 1 even in regular row echelon form.