Does Cramer's Rule give fake solutions?

Question

$$\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \\ \end{bmatrix}\times\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \\ \end{bmatrix}$$ If the system of given linear equations, has infinitely many solutions, then find $\alpha$.

I use the Cramer's Rule, like the way it is used here.

So I define $\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \\ \end{bmatrix}= \Delta$ which should be equal to zero (Cramer's Rule). Solving this I get two values of $\alpha$ : 1 and -1 .

Now we proceed to calculate $\Delta_1$,$\Delta_2$,$\Delta_3$ as follows:

$\Delta_1=\begin{bmatrix} 1 & \alpha & \alpha^2 \\ -1 & 1 & \alpha \\ 1 & \alpha & 1 \\ \end{bmatrix}$
$\Delta_2=\begin{bmatrix} 1 & 1 & \alpha^2 \\ \alpha & -1 & \alpha \\ \alpha^2 & 1 & 1 \\ \end{bmatrix}$
$\Delta_3=\begin{bmatrix} 1 & \alpha & 1 \\ \alpha & 1 & -1 \\ \alpha^2 & \alpha & 1 \\ \end{bmatrix}$

If I put $ \alpha = 1 $ and $\alpha = -1$ in the above matrices, we get all the determinants equal to zero ($\Delta_1=\Delta_2=\Delta_3=0$). So I conclude both 1 and -1 should be a solution to the question.

Now the problem arises as I inspect the answer. The equations formed when I put $\alpha = 1$ will be:
$x+y+z=1 \\ x+y+z=-1 \\ x+y+z=1$

Here the first two planes are parallel to each other. That means they donot intersect and must imply inconsistency. So I don't have infinite solutions for $\alpha = 1$. So the sole answer would be $\alpha = -1$ .

My Question is why does $\alpha = 1$ satisfy the equations but does not satisfy the problem. Why does solving give $\alpha=1$ when it isn't a solution?

And do I need to do this kind of inspection for every equation that I solve with this method? (Since it tends to give superfluous answers).

All help is appreciated.

EDIT:

I thought it would be best if I clarified my question a little bit more.

Cramer's Rule tells us that :

For a Unique Solution $\Delta \ne 0$ (we don't care about the other deltas) And the solutions will be $\Delta_1/\Delta$ $\Delta_2/\Delta$ , $\Delta_3/\Delta$

For no solution $\Delta = 0$ and atleast one of the other Deltas is non-zero.

For infinite solutions $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$.

Now since $\Delta$ in the question is zero, there are no unique solutions. That leaves us with two options: either no-solution or infinite-solutions.

So equating all the other Deltas to zero simultaneously I get $\alpha = 1,-1$ as stated before.

But then why does $\alpha = 1$ give me no solutions.

I few minutes of stroll below and you'll see that this method isn't that popular (especially the infinite one). I don't know why. But I'm sure somebody out there will definitely follow on.

Answer 1

The Cramer's Rule says that if the system has only one solution ($\Delta \ne0$), then solution is given by,

$$x_i=\frac{\Delta x_i}{\Delta}$$

So if $\Delta=0$ then you can't use Cramer's Rule.

In general you have that $(\Delta)x_i=\Delta x_i$ so, if $\Delta=0$ and $\Delta x_i=0$ you can't say anything about the solution.

Answer 2

Your determinant should not be $0$ because you want to divide other determinants by it. $$\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \\ \end{bmatrix} = (\alpha ^2 -1)(\alpha ^3 -1)$$ Find other deerminants and divide to get your answers in terms of $\alpha$

Answer 3

It is something wrong with the source you are using. It says

The system is dependent if all the determinants have a value of zero.

It is wrong. A simple counterexample: the system $$ \begin{bmatrix} 0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} 1\\1\end{bmatrix} $$ is obviously inconsistent, however, all the determinants are zero $$ \Delta=0,\quad\Delta_x=\begin{vmatrix}1 & 0\\1 & 0\end{vmatrix}=0,\quad\Delta_y=\begin{vmatrix}0 & 1\\0 & 1\end{vmatrix}=0. $$ Another example: $$ \begin{bmatrix} \alpha & 0\\0 & \alpha\end{bmatrix}\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} 1\\1\end{bmatrix}. $$ Here $$ \Delta=\alpha^2,\quad \Delta_x=\Delta_y=\alpha. $$ For $\alpha=0$ we have as above all the determinants are zero, however, the calculation of e.g. $x$ gives $$ x=\frac{\Delta_x}{\Delta}=\frac{\alpha}{\alpha^2}. $$ When $\alpha\to 0$ we get that the zero in the denominator is "more zero" then in the numerator, so we get $\frac{1}{\alpha}$ not existing limit.