# Matrix method of solving non homogenous linear equations

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.

• Did you not make a mistake in calculating the adjoint? – Gerry Myerson Aug 14 at 7:43
• I don't think so. I got $adj(A)=\left[ \begin{array}{ccc} - \alpha^{2} + 1 & \alpha^{3} - \alpha & 0 \\\\ \alpha^{3} - \alpha & - \alpha^{4} + 1 & \alpha^{3} - \alpha \\\\ 0 & \alpha^{3} - \alpha & - \alpha^{2} + 1 \end{array} \right] – thewitness Aug 14 at 8:12 • I verified this with calculator also. And this is a null matrix for both$\alpha = \pm 1$– thewitness Aug 14 at 8:13 • My apologies, you're quite right. Well, then, something must be wrong with the statement that if$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. – Gerry Myerson Aug 14 at 12:45
• I'm voting to close this question as off-topic because the question is incorrect – thewitness 2 days ago