I've been given the following system of equations:

$$
(4-\lambda)x\_1-2x\_2-x\_3=1\\
-2x\_1+(1-\lambda)x\_2+2x\_3=2\\
-x\_1+2x\_2+(4-\lambda)x\_3=1
$$

The resulting coefficient matrix would be:

$$
\begin{bmatrix}
(4-\lambda)&-2&\ &-1&\ &1\\\\
-2&(1-\lambda)&\ &2&\ &2\\\\
-1&2&\ &(4-\lambda)&\ &1
\end{bmatrix}
$$

I've managed to reduce it to:

$$
\begin{bmatrix}
(4-\lambda)&-2&-1&\ &1\\\\
0&\frac{-6-\lambda}{4-\lambda}&(5-\lambda)&\ &\frac{3-\lambda}{4-\lambda}\\\\
0&(5-\lambda)&(\lambda-5)&\ &0
\end{bmatrix}
$$

There are two claims I'm being asked to answer either true or false on:

1. If $\lambda\neq -1,5$ the system of equations has only one solution
2. The system of equations has a solution for any value of $\lambda$

Is it possible to confirm or deny these claims without actually solving the system? Because I've honestly tried reducing it further and gotten a pretty messy resulting matrix. Can someone shed some light on this for me please?