Hi MHF,
Here is an exercise and I've done part a) and b) but not part c). So I'd like a check up for the 2 first parts and a tip for the last part. Thanks in advance.
Consider the system of equations given by $\displaystyle w-2x-y+3z=a$, $\displaystyle -w+x-2y+2z=b$, $\displaystyle 2w-3x+y+z=c$.
a)Show that the system has at least 2 solutions that are linear independent if $\displaystyle a=b=c=0$.
My attempt : I wrote the system of equations as a matrix and reduced it in order to get $\displaystyle \begin{bmatrix} 1 & 0 & 5 & -7 \\ 0 & 1 & 3 & -5 \\ 0 & 0 & 0 & 0 \end{bmatrix}$. From which we get $\displaystyle \begin{cases} w+5y-7z=0 \\ x+3y-5z=0 \end{cases}$ so any vector in the plane can be written as $\displaystyle y(-5, -3, 1, 0)+z(7, 5, 0, 1)$. Let $\displaystyle y=1$ and $\displaystyle z=0$, we get that a solution is $\displaystyle \{ (-5, -3, 1, 0 ) \}$, let $\displaystyle y=0$ and $\displaystyle z=1$ and we get $\displaystyle \{ (7, 5, 0, 1 ) \}$ which are clearly 2 linear independent solutions.

b)Determine the values of a, b and c such that the system has solution.
My attempt : I wrote the system as an augmented matrix (augmented with the column a, b, c) and finally reaches the same matrix as before of course, but with the column vector $\displaystyle (2a+b, -b-a, c-2a+b+a)$ from which we have $\displaystyle \begin{cases} w+5y-7z=2a+b \\ x+3y-5z=-b-a \\ 0=c-2a+b+a \end{cases}$ so $\displaystyle w+x+8y-12z=a$, $\displaystyle b$ is any value and $\displaystyle c=a-b$.

c)Show all the solutions for $\displaystyle a=1$, $\displaystyle b=0$ and $\displaystyle c=1$, if there are/is.
My attempt : $\displaystyle \begin{cases} w+5y-7z=2 \\ x+3y-5z=-1 \end{cases}\Rightarrow w+x+8y-12z=1$ so the solutions are infinitely many? All numbers $\displaystyle w$,$\displaystyle x$,$\displaystyle y$ and $\displaystyle z \in \mathbb{R}$ such that $\displaystyle w+x+8y-12z=1$ are solutions?