Find constants in system of equations In the below system of equations, find the values of constants $q_1, q_2, q_3 \in \mathbb R$ so that the system has no solution.
$$\begin{cases}
2x&+y-z&+2s &= q_1\\
3x&+y-2z&+s &= q_2\\
x&+y+3z&-s &= q_3
\end{cases}$$
In order to eliminate some of the unknowns, we add the 1st with the 3rd and subtract the 2nd, so we get:
$y+4z = q1+q3-q2$
By similar more ways (multiplying, adding and subtracting) we get some more such equations.
I don't see how this system can have no solution; plus that we are one equation short (3 equations - 4 unknowns).
Any assistance is much appreciated.
EDIT:
(Apologies but I am not familiar with linear algebra!!)
What I believe I must do is the following:
Reduce the unknowns by one (don't see how), so as to have a 3x3 system. Calculate the determinant, for which it must be $D \neq 0$.
Then calculate $D_x, D_y, D_z$ assuming we have eliminated s.
Then, for the system NOT to have a solution, it must be $D=0$ and $D_x, D,y, D_z \neq 0$.
Any further help?
 A: The system can be written as$$\begin{pmatrix}
2& 1 &-1& 2\\
3&1 &-2& 1\\
1&1& 3&-1\end{pmatrix} \begin{pmatrix}
x\\ y \\ z \\ s\end{pmatrix} = \begin{pmatrix}
q_1\\ q_2 \\ q_3\end{pmatrix}.$$ Now since we have that $$\mathrm{rank}\begin{pmatrix}
2& 1 &-1& 2\\
3&1 &-2& 1\\
1&1& 3&-1\end{pmatrix} =3$$ the system must be consistent, independently of the values of $q_1,q_2,q_3$. (Obviously it is $$\mathrm{rank}\begin{pmatrix}
2& 1 &-1& 2\\
3&1 &-2& 1\\
1&1& 3&-1\end{pmatrix} =\mathrm{rank}\begin{pmatrix}
2& 1 &-1& 2 &q_1\\
3&1 &-2& 1 & q_2\\
1&1& 3&-1 & q_3\end{pmatrix}=3.)$$
EDIT
$$\begin{cases}
2x&+y&-z&+2s &= q_1\\
3x&+y&-2z&+s &= q_2\\
x&+y&+3z&-s &= q_3
\end{cases} \iff \begin{cases}
2x&+y&-z &= q_1-2s\\
3x&+y&-2z&= q_2-s\\
x&+y&+3z&= q_3+s
\end{cases} $$
$$\iff  \begin{cases}
2x&+y&-z &= q_1-2s\\
x&&-z&= q_2-q_1+s\\
-x&&+4z&= q_3-q_1+3s
\end{cases}$$
$$\iff  \begin{cases}
2x&+y&-z &= q_1-2s\\
x&&-z&= q_2-q_1+s\\
&&3z&= q_3+q_2-2q_1+4s
\end{cases}$$
From the last equation we get
$$z=\dfrac{q_3+q_2-2q_1+4s}{3}.$$
Now $$x=z+q_2-q_1+s=\dfrac{q_3+4q_2-5q_1+7s}{3}.$$
Finally we get $$y=z-2x+q_1-2s=\dfrac{-q_3-7q_2+11q_1-16s}{3}.$$ That is, the system is consistent independently of the values of $q_1,q_2,q_3.$
A: Thanks to Rouché-Capelli theorem we can state that the given system has always solutions.
Indeed the rank of the matrix $A$ of the coefficients is $3$ and the augmented matrix $A|B$ can't have a different rank, even if $q_i$ were all zero.
