Let $a,b,c$ are distinct integers. Then which among the following is the set of possible rank of $A$=

Question

Let $a,b,c$ are distinct integers. Then which among the following is the set of possible rank of $$A = \begin{pmatrix} a^2 & b^2 & c^2 \\ a^5 & b^5 & c^5 \\ a^{11} & b^{11} & c^{11} \end{pmatrix} $$

1. $\{0,1,2,3\}$

2. $\{1,2,3\}$

3. $\{2,3\}$

4. $\{3\}$

My Attempt:

$$A = \begin{pmatrix} a^2 & b^2 & c^2 \\ a^5 & b^5 & c^5 \\ a^{11} & b^{11} & c^{11} \end{pmatrix} $$

Let $$det(A) = det\begin{pmatrix} a^2 & b^2 & c^2 \\ a^5 & b^5 & c^5 \\ a^{11} & b^{11} & c^{11} \end{pmatrix} $$

Taking common $a^2,b^2,c^2$ from $C_1,C_2,C_3$ respectively, we get

$$det(A) = a^2b^2c^2. det\begin{pmatrix} 1 & 1 & 1 \\ a^3 & b^3 & c^3 \\ a^9 & b^9 & c^9 \end{pmatrix} $$

Operating $C_1 \to C_1 - C_2$,$C_2 \to C_2 - C_3$, we get

$$det(A) = a^2b^2c^2. det\begin{pmatrix} 0 & 0 & 1 \\ a^3-b^3 & b^3-c^3 & c^3 \\ a^9-b^9 & b^9-c^9 & c^9 \end{pmatrix} $$ $$det(A) = a^2b^2c^2. det\begin{pmatrix} 0 & 0 & 1 \\ a^3-b^3 & b^3-c^3 & c^3 \\ (a^3-b^3)(a^6+b^6+a^3b^3) &(b^3-c^3)(b^6+c^6+b^3c^3) & c^3 \end{pmatrix} $$

Taking $(a^3-b^3),(b^3-c^3)$ common from $C_1,C_2$ respectively, we get

$$det(A) = a^2b^2c^2(a^3-b^3)(b^3-c^3). det\begin{pmatrix} 0 & 0 & 1 \\ 1 & 1 & c^3 \\ (a^6+b^6+a^3b^3) &(b^6+c^6+b^3c^3) & c^3 \end{pmatrix} $$

Expanding by $R_1$, we get $det(A) = a^2b^2c^2(a^3-b^3)(b^3-c^3)[-(a^6+b^6+a^3b^3)+(b^6+c^6+b^3c^3)] = a^2b^2c^2(a^3-b^3)(b^3-c^3)[(c^6-a^6)+b^3(c^3-a^3)] = a^2b^2c^2(a^3-b^3)(b^3-c^3)[((c^3)^2-(a^3)^2)+b^3(c^3-a^3)] = a^2b^2c^2(a^3-b^3)(b^3-c^3)[(c^3-a^3)(c^3+a^3)+b^3(c^3-a^3)] = a^2b^2c^2(a^3-b^3)(b^3-c^3)(c^3-a^3)(a^3+b^3+c^3)$

If $a \neq b \neq c$ then Rank of $A$ may be $3$. I could not determine other possibility of ranks. Please help me.

Answer 1

As you noted there are examples where the rank is $3$, for an example where the rank is $2$ take $a=0,b=1,c=-1$ or whatever.

The rank cannot be $0$ clearly.

The rank cannot be $1$ either, assume the rank is $1$. Without loss of generality both $a$ and $b$ are different to $0$, it follows that if the rank is $1$ then the second column must be a scalar multiple of the first,let $\lambda$ be the scalar, then $\lambda = a^2/b^2$ and $\lambda = a^5/b^5$ and so $(a/b)^3 = 1 \implies a/b = 1$, contradicting $a\neq b$.