Find trivial and non-trivial solutions to a system with a parameter

Question

\begin{cases} x + (2p−1)y−pz = 0 \\ (2−p)x + y−z = 0 \\ x + py−z = 0 \end{cases}

• 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

Answer 1

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.

Answer 2

Hint :

Apply Gauss-Elimination to :

$$\left(\begin{array}{ccc|c} 1 & 2p-1 & -p &0 \\ 2-p & 1 & -1 & 0 \\ 1 & p & -1 & 0\end{array}\right)$$

Answer 3

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}