Find trivial and non-trivial solutions to a system with a parameter \begin{cases}   x + (2p−1)y−pz = 0 \\
    (2−p)x + y−z = 0 \\
    x + py−z = 0 \end{cases}


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*For which values of $p$, system has trivial solutions?

*For which values of $p$, system has nontrivial solutions? Write them


Now I have come to $p_1=1, p_2=1, p_3=1/2$ with Sarrus rule. Now how do I get trivial and nontrivial solutions ? I am just looking for instructions 
 A: Hint :
Apply Gauss-Elimination to :
$$\left(\begin{array}{ccc|c} 1 & 2p-1 & -p  &0 \\ 2-p  & 1 & -1 & 0 \\ 1 & p & -1 & 0\end{array}\right)$$
A: Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.
So the answer to the first is: For all values of $p$ there will be trivial solutions.
It the equations are linear dependent then those are the only solutions.  But if they aren't linearly independent they can have non-trivial solutions.  Which can only occur if the determinate is $0$.  So this question boils down to nothing more or less than:  For what values of $p$ will the determinate be $0$.  (Note: there is nothing in the question about finding the solutions).
Using rule of sarrus the determinate is:
$-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$
$ p^3 - 4p^2  +5p-2= (p-1)^2(p-2)$
Which if we set to $0$ and solve... we get
$p =1$ or $p =2$.
And $p = 1$ we get the system:
$x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.
And if $p=2$ we get the system:
$x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.
Bear in mind $\begin{cases}   a_1x + a_2y+a_3z = 0 \\
    b_1x + b_2y+b_3z = 0 \\
    c_1x + c_2y+c_3z = 0 \end{cases}$
Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.
A: Hint:
This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
\begin{align}
\begin{vmatrix}
1&2p-1&-p \\ 2-p&1&-1 \\ 1&p&-1
\end{vmatrix} &=
\begin{vmatrix}
1&2p-1&-p \\ 0&1+(p-2)(2p-1)&-1-p(p-2) \\ 0&1-p&p-1
\end{vmatrix} 
=\begin{vmatrix}
1&p-1&-p \\ 0&(p-2)(p-1)&-1-p(p-2) \\ 0&0&p-1
\end{vmatrix}\\[1ex]& = (p-1)\begin{vmatrix}
1& 1&-p \\ 0&p-2 &-1-p(p-2) \\0&0&p-1
\end{vmatrix} =  (p-1)\begin{vmatrix}
 p-2 &-1-p(p-2) \\ 0&p-1
\end{vmatrix} \\[1ex]
&=(p-1)^2(p-2).
\end{align}
