Augmented Matrix For which real numbers $ a$ and $b$ does the following linear system have 
(a) no solution,
(b) exactly one solution, or 
(c) infinitely many solutions? Justify your answers. 
$$ (a-1)x + (a+3)y + z = 1 $$
$$ax + a^2y + z = 1$$
$$x + ay + z  = b$$
i start by putting it as an augmented matrix, but then i dont know how to reduce it into row echelon form or reduced ref. because of the algebraic terms. is there anyone that can help? thanks a lot...
 A: $$\begin{cases}(a-1)x + (a+3)y + z = 1 \\
ax + a^2y + z = 1 \\
x + ay + z = b\end{cases} \iff \left[\begin{array}{ccc|c} a-1 & a+3 & 1 & 1 \\ a & a^2 & 1 & 1 \\ 1 & a & 1 & b \end{array}\right]$$
OK, now we just do Gauss-Jordan:
$$\left[\begin{array}{ccc|c} a-1 & a+3 & 1 & 1 \\ a & a^2 & 1 & 1 \\ 1 & a & 1 & b \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & a & 1 & b \\ a & a^2 & 1 & 1 \\ a-1 & a+3 & 1 & 1 \end{array}\right]\\ \sim \left[\begin{array}{ccc|c} 1 & a & 1 & b \\ 0 & 0 & 1-a & 1-ab \\ 0 & a+3-(a^2-a) & 1-(a-1) & 1-b(a-1) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & a & 1 & b \\ 0 & 0 & -a+1 & -ab+1 \\ 0 & -a^2 + 2a+3 & -a+2 & -ab+b+1 \end{array}\right] \\ \sim \left[\begin{array}{ccc|c} 1 & a & 1 & b \\ 0 & -a^2 + 2a+3 & -a+2 & -ab+b+1 \\ 0 & 0 & -a+1 & -ab+1 \end{array}\right] \\ \sim \left[\begin{array}{ccc|c} 1 & a & 1 & b \\ 0 & 1 & \frac{a-2}{(a+1)(a-3)} & \frac{ab-b-1}{(a+1)(a-3)} \\ 0 & 0 & 1 & \frac{ab-1}{a-1} \end{array}\right]$$
Be sure to keep track of anything you divide by -- for instance I divided by $-a^2+2a+3=-(a+1)(a-3)$ in the second row of the last step and $-a+1$ in the last row of the last step.  So you'll need to try $a=-1$, $a=3$, and $a=1$ individually.
Does that help?
