Conclude that $A$ is invertible if and only if $\alpha, \beta$ and $\gamma$ are pairwise different

Question

Let $\alpha, \beta$ and $\gamma$ be real numbers and consider the matrix $$ A = \begin{pmatrix} 1 & \alpha & \alpha ^2 \\ 1 & \beta & \beta^2 \\ 1 & \gamma & \gamma^2 \end{pmatrix} \in \ \text{Mat}_3(\mathbb{R}) $$ I have to conclude that $A$ is invertible if and only if $\alpha, \beta$ and $\gamma$ is pairwise different (I hope this is the right translation as I have translated from Danish into English).

My attempt:

First way: Let $A$ be an invertible matrix. As $A$ is invertible we have that $\det(A) \neq 0 $. But this is only the case if $\alpha, \beta$ and $\gamma$ is pairwise different as if they were pairwise equal $\det(A) = 0$

Second way: Assume $\alpha, \beta$ and $\gamma$is pairwise different. If this is the case we must have $\det(A) \neq 0$ as if $\alpha, \beta$ and $\gamma$ were pairwise equal we would have that $\det(A) = 0$. But if $\det(A) \neq 0$ we must have that A is an invertible matrix.

Thus we conclude that $A$ is invertible if and only if $\alpha, \beta$ and $\gamma$ is pairwise different.

Is this approach alright? Can you verify? Thanks in advance!

Answer 1

No, it is not correct. Yes, if $\alpha$, $\beta$ and $\gamma$ are not parwise distinct, then $\det(A)=0$. But you cannot deduce from that that otherwise $\det(A)\ne0$.

You can solve your problem by proving that $\det(A)=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$.

Answer 2

I think you're overcomplicating the argument.

• $A$ is invertible iff $\det A \ne 0$

• $\det A = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$. That's the main point.

• $A$ is invertible iff $(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha) \ne 0$ iff $\alpha$, $\beta$, $\gamma$ are pairwise different.