Analyze solutions to a matrix 3x5 matrix with two parameters and find a unique solution I have the linear system (over $\mathbb{R}$) for which I need to find a unique solution:
$$\begin{cases}
4x+8y+7z+3cw = 3b \\
x+2y+2z+cw=b \\
2x+4y+2z+(c-1)w=b
\end{cases}$$
for which the corresponding matrix is:
$$\begin{bmatrix}
4&8&7&3c&3b \\
1&2&2&c&b \\
2&4&2&c-1&b
\end{bmatrix}
$$
After the following elementary operations:
$R_2 \to R_1; R_2 \to R_2-4R_1, R_3 \to R_3-2R1;R_3 \to R_3-2R_2; R_2 \to -R_2$
I managed to get to the reduced form:
$$\begin{bmatrix}
1&2&2&c&b \\
0&0&1&c&b \\
0&0&0&c-1&b
\end{bmatrix}
$$
At this point I can see which conditions need to hold for the matrix:
1) to not have any solution: $c=1, b \neq 0$
2) to have infinitely many solutions: $c=1, b=0$ or $c \neq 1, b=0$ or $c \neq 1, b \neq 0$
3) $\mathbf{However}$ I don't see any way how there'd be a unique solution to the system because we have more variables than equations and the column of $w$ variable cannot be a column of zeroes.
 A: So we have system of equations:
$\begin{cases}
4x+8y+7z+3cw=3b
\\
x+2y+2z+cw=b
\\
2x+4y+2z+(c-1)w=b
\end{cases}$
And the matrix of it is:
$\begin{bmatrix}
4&8&7&3c&3b
\\
1&2&2&c&b
\\
2&4&2&c-1&b
\end{bmatrix}$
Using these steps:
$1)\,R1\rightleftharpoons R2$
$2)\,R2=R2-4R1$
$3)\,R3=R3-2R1$
$4)\,R1=R1+R2$
$5)\,R3=R3-2R2$
$6)\,R2=(-1)\times R2$
I got:
$\begin{bmatrix}
1&2&1&0&0
\\
0&0&1&c&b
\\
0&0&0&c-1&b
\end{bmatrix}$
$\Rightarrow\,
\begin{cases}
x+2y+z=0
\\
z+cw=b
\\
(c-1)w=b
\\
y\quad is\,free
\end{cases}$
Then I started solving them:
$1)\quad (c-1)w=b \\ \hspace{9mm}w=\frac{b}{c-1}$
$2)\quad z+cw=b \\
\hspace{9mm}z+c\times\frac{b}{c-1}=b \\
\hspace{9mm}z+\frac{cb}{c-1}=b \\
\hspace{9mm}z\times (c-1)+cb=b\times (c-1) \\
\hspace{9mm}z(c-1)+cb=cb-b \\
\hspace{9mm}z(c-1)=cb-b-cb \\
\hspace{9mm}z(c-1)=-b \\
\hspace{9mm}z=-\frac{b}{c-1} $
$3)\quad x+2y+z=0 \\
\hspace{9mm}x=-2y-z \\
\hspace{9mm}x=-2y-(-b) \\
\hspace{9mm}x=b-2y$
So the unique solution is:
$\begin{cases}
x=b-2y
\\
y=y
\\
z=-\frac{b}{c-1}
\\
w=\frac{b}{c-1}\hspace{18mm},\,c-1\neq 0
\end{cases}$
