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For a real number $\alpha$, if the system of linear equations $\begin{bmatrix} 1 & \alpha & \alpha ^{2} \\ \alpha & 1 & \alpha \\ \alpha ^{2} & \alpha & 1 \end{bmatrix}$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ $ = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$ has infinitely many solutions, find $\alpha$.

Let

$A=\begin{bmatrix} 1 & \alpha & \alpha ^{2} \\ \alpha & 1 & \alpha \\ \alpha ^{2} & \alpha & 1 \end{bmatrix}$

$X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$

$D=\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$

The equation can be written as $AX=D$

I learnt that this system(for $D \neq 0$. i.e. non homogenous system of equations):

has a unique solution if $A$ is a non-singular matrix($|A| \neq 0$)

has infinite number of solutions if $A$ is a singular matrix ($|A|=0$) and $adj(A)*D=0$.

has no solution if $A$ is a singular matrix ($|A|=0$) and $adj(A)*D \neq 0$.

Solving for $|A|=0$, we get $\alpha = \pm 1$.

But here comes my doubt. In this case, $adj(A)=0$ (null matrix) for both $\alpha = \pm 1$. So $adj(A)*D=0$ for both those values. So according to what I learnt, for both $\alpha = \pm 1$ there should be infinite solutions to this system of equations.

But on expanding the matrix into a system of linear equations, it is easily observed that for $\alpha=1$ there are no solutions and for $\alpha=-1$ there are infinite solutions.

I feel that the condition $adj(A)*D$ equal to or not equal to 0 for infinite or no solution respectively is valid only if $adj(A) \neq 0$. Is this correct?

How can we deal with this case without having to expand it into a system of linear equations and verifying with values? (as that can get very tedious at times)

P.S: I have verified all my calculations with a calculator as well so there are no calculation mistakes.

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  • $\begingroup$ Did you not make a mistake in calculating the adjoint? $\endgroup$ – Gerry Myerson Aug 14 at 7:43
  • $\begingroup$ I don't think so. I got $adj(A)=\left[ \begin{array}{ccc} - \alpha^{2} + 1 & \alpha^{3} - \alpha & 0 \\\\ \alpha^{3} - \alpha & - \alpha^{4} + 1 & \alpha^{3} - \alpha \\\\ 0 & \alpha^{3} - \alpha & - \alpha^{2} + 1 \end{array} \right] $\endgroup$ – thewitness Aug 14 at 8:12
  • $\begingroup$ I verified this with calculator also. And this is a null matrix for both $\alpha = \pm 1$ $\endgroup$ – thewitness Aug 14 at 8:13
  • $\begingroup$ My apologies, you're quite right. Well, then, something must be wrong with the statement that if $A$ is singular, and that product is zero, then there are infinitely many solutions. Do you have a citation for that? Have a look at math.stackexchange.com/questions/2653517/… and see whether that helps. $\endgroup$ – Gerry Myerson Aug 14 at 12:45
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    $\begingroup$ I'm voting to close this question as off-topic because the question is incorrect $\endgroup$ – thewitness 2 days ago

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