$$\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \\ \end{bmatrix}\times\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \\ \end{bmatrix}$$ 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.


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.

  • 1
    $\begingroup$ Why do you set $\Delta$ to be equal to zero as your first step? $\endgroup$ – MSobak May 9 '18 at 18:43
  • 1
    $\begingroup$ Look at this section, especially about the systems with three or more equations: en.wikipedia.org/wiki/… $\endgroup$ – Andrei May 9 '18 at 18:43
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    $\begingroup$ @Sobi for infinite solutions $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$ $\endgroup$ – SmarthBansal May 9 '18 at 18:45
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    $\begingroup$ You obtain Cramer's rule by multiplying the system by the adjugate matrix of the system. If the determinant of the system is zero, then this is akind of multiplying by $0$ (strictly speaking a matrix that is a zero divisor got multiplied by a matrix that annihilates it). Notice that an incompatible equation $0=1$ can be turn compatible by multiplying by $0$, since $0\times 0=0\times 1$ is compatible. That is why Cramer's rule gives you candidate systems with infinite solutions. You then need to check directly if the original system was compatible or not. $\endgroup$ – user551819 May 9 '18 at 18:57
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    $\begingroup$ People are pointing out that Cramer's rule doesn't say what you want. The answer is thus: no, Cramer's rule isn't wrong; you are using the wrong rule, and it doesn't really matter why you're using it. The page to which you link says "If the denominator determinant, $D$ [your $\Delta$], has a value of zero, then the system is either inconsistent or dependent." This is the phenomenon you have observed: if the system is underdetermined, then $\Delta = 0$; but one cannot conclude the converse, and, in fact, a system with $\Delta = 0$ may be inconsistent. $\endgroup$ – LSpice May 9 '18 at 18:58

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 $\alpha$

  • $\begingroup$ I specifically want infinite solutions so $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$. should hold. Thanks for writing this answer +1 $\endgroup$ – SmarthBansal May 9 '18 at 18:55

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.

  • $\begingroup$ I specifically want infinite solutions so $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$ Thanks for the answer though +1 $\endgroup$ – SmarthBansal May 9 '18 at 18:56
  • $\begingroup$ @SmarthBansal: In your case you have to use Gauss elimination or something equivalent $\endgroup$ – Arnaldo May 9 '18 at 18:58
  • $\begingroup$ Though I'm not sure why my method fails. But I will read about Gauss elimination as you say. $\endgroup$ – SmarthBansal May 9 '18 at 18:59
  • $\begingroup$ @SmarthBansal: Your method fails because he is not valid when $\Delta =0$. $\endgroup$ – Arnaldo May 9 '18 at 19:01

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.

  • $\begingroup$ +1 Thanks for the answer! I agree your examples do support your answer. But this isn't the only place where I've seen this sort of method. And it does work well for many other equations that I've seen........ $\endgroup$ – SmarthBansal May 9 '18 at 20:12
  • $\begingroup$ ........For example: $2x + Py + 6z = 8$; $x+2y+Qz = 5$; $x+y+3z = 4$ ; Here if you solve for $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$ you get $P=2$, $Q=3$ which does really cause the equations to have infinite solutions. $\endgroup$ – SmarthBansal May 9 '18 at 20:13
  • $\begingroup$ Do you think this method might be missing some boundaries, which invalidates the examples you mentioned and many more alike? $\endgroup$ – SmarthBansal May 9 '18 at 20:13
  • $\begingroup$ @SmarthBansal The method cannot be used when all the determinants are zeros. The standard solution is to calculate those exceptional zeros where $\Delta=0$ and then for each particular exceptional value perform elimination, i.e. solve the linear system. $\endgroup$ – A.Γ. May 9 '18 at 20:34

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