When is a nonhomogeneous linear system solvable I'm supposed to determine for which t the linear system $\bf{A}*x = y$ is solvable 
$\bf{A}$ = $\begin{bmatrix}1 & 0 & 1\\0 & -1 & 1\\ t-2 & 0 &0 \end{bmatrix},\quad \bf{y}  = \begin{bmatrix} t-1 \\ 1 \\ 1 \end{bmatrix}$.
My thought was that  $t-2$ must be equal to $1$, therefore the linear system can only have an answer for $t=3$. 
So then I plug in $3$ for every $t$ and can carry on as usual(?)
 A: Hint:
The criterion for a non-nomogeneous linear system to have solutions is that the matrix $A$ and the augmented matrix $[A|\mathbf y]$ have the same rank. So proceed by Gaußian elimination.
A: It isn't true that $t-2$ must be equal to $1.$ Rather, we would have $(t-2)x_1=1.$ However, this does tell us that we must have $t-2\neq 0,$ so we can't allow $t=2$ if we want the system to be solvable.
Assuming that $t\neq 2,$ we can see that $A$ row-reduces to the identity matrix (or, more simply, that $A$ has non-zero determinant), meaning that $A$ is invertible. Thus, the system is solvable when (and only when) $t\ne 2.$
A: Consider the row reduction
$$
\left[\begin{array}{rrr|r}
1 & 0 & 1 & t - 1 \\
0 & -1 & 1 & 1 \\
t - 2 & 0 & 0 & 1
\end{array}\right]
\xrightarrow{R_3-(t-2)\cdot R_1\to R_3}
\left[\begin{array}{rrr|r}
1 & 0 & 1 & t - 1 \\
0 & -1 & 1 & 1 \\
0 & 0 & -t + 2 & -t^{2} + 3 \, t - 1
\end{array}\right]
$$
This tells us that the rank of the coefficient matrix is three if $t\neq 2$ and two if $t=2$. We can then proceed in cases.
Suppose $t=2$. Then our above system is
$$
\left[\begin{array}{rrr|r}
1 & 0 & 1 & 1 \\
0 & -1 & 1 & 1 \\
0 & 0 & 0 & 1
\end{array}\right]
$$
This system is unsolvable, since the last equation is $0=1$ which is impossible.
Can you sort out what happens if $t\neq 2$?
