Confused about a linear equation

Question

So I am working through some notes on Linear Algebra and I cant seem to follow this one part. The question asks to Solve:

$x+y-z+2w=-20$

$2x-y+z+w=11$

$3x-2y+z-2w+27$

I don't have a problem with putting the equation into matrix form and even reducing it. The way the notes explains it is what I don't understand at all. So first once its in echelon, $$ \begin{bmatrix} 1 & 1 & -1 & 2 & -20\\ 0 & 1 & -1 & 1 & -17\\ 0 & 0 & 1 & 3 &-2\\ \end{bmatrix} $$the notes states: "The solution is therefore

$$ X=\begin{pmatrix} -w-3\\ -4w-19\\ -3w-2\\ 2\\ \end{pmatrix} $$

(This is the first place I am confused? Why are there $4$ rows now? And how did they get these numbers?).

Then, we can reduce and have $$ \begin{bmatrix} 1&0&0&1&-3\\ 0&1&0&4&-19\\ 0&0&1&3&-2 \end{bmatrix} $$

Now for the second confusing part, it then states: "Note, the solution is

$$ X=\begin{pmatrix} -3\\ -19\\ -2\\ 0\\ \end{pmatrix} +w\begin{pmatrix} -1\\ -4\\ -3\\ 1\\ \end{pmatrix} $$

Note that the rank is $3$ and there is $1$ parameter.

Thanks a lot for the help guys. Hopefully I can understand it.

Answer 1

You have four unknowns, so a solution is a vector with four components.

The system, after the final reduction, can be written as $$ \begin{cases} x+w=-3\\ y+4w=-19\\ z+3w=-2 \end{cases} $$ Since $w$ is a free variable, you can set it to any value you want; therefore the solutions are $$ \begin{cases} x=-3-w\\ y=-19-4w\\ z=-2-3w\\ w=w \end{cases} $$ that in vector form can be written as $$ \begin{bmatrix} -3-w\\ -19-4w\\ -2-3w\\ w \end{bmatrix}= \begin{bmatrix} -3\\ -19\\ -2\\ 0 \end{bmatrix}+ w\begin{bmatrix} -1\\ -4\\ -3\\ 1 \end{bmatrix} $$ where $w$ is any number.

Answer 2

Assuming your columns are x,y,z,w, your echelon form translates to

z + 3w - -2 or z = 2- 3w, the next row up says y + 4 w = -10 or y = -10 -4w, continuing x + w = -3, So solutions look like they have a freely chosen w from which says solutions look like: (x, y,z,w) = (-3-w, -10-4w, 2-3w, w) or w(-1,-4,-3 1) + (-3, -10, 2,0), a one-parameter family solutions with parameter w. Of course, you could write it as a column just as well. The solution is always of this form, an element of the kernel* + a particular solution (Here, the particular solution is (x,y,z,w) = ( -3, -19, -2, 0).

*The kernel is the set of vectors which are mapped to zero by the original matrix.

Because you have infinitely many solutions the original matrix has det = 0. You should't have been surprised that your solutions are 4-tuples, because you have 4 unknowns. This is actually incomplete because I do not know the precise statement of the original problem, I.E. Was it a homogeneous system? That requires a slightly different interpretation.

Hope this helps!