How do I eliminate the repeating cases?

Question

Let $K$ be the field with exactly $7$ elements. Let $\mathscr M$ be the set of all $2×2$ matrices with entries in $K$. How many elements of $\mathscr M$ are similar to the following matrix?

$ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$

My attempt: Answer is given is $56$. We need to find the cardinality of $\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} a,b,c,d\in K:a+d=1 \wedge ad=bc\}=\{\begin{pmatrix} a & b \\ c & 1-a \end{pmatrix} a,b,c,d\in K: ad=bc\}.$ I got $ad=bc \implies a(1-a)=bc \implies a-a^2=bc\implies $ $a$ as a function of $bc$. So, there are $7^3$ possibilities.How do I eliminate the repeating cases?

Answer 1

You can continue as below: $$a-a^2=bc \Longrightarrow a^2-a+bc=0 \Longrightarrow a=\frac{1\pm\sqrt{1-4bc}}{2}\overset{1/2 = 4}{===}4\pm4\sqrt{1-4bc}$$ So $(1-4bc)$ must be square: $$(1-4bc) \in \{ 0^2,(\pm)1^2,(\pm2)^2,(\pm3)^2 \} = \{ 0,1,4,2 \} \Longrightarrow 4bc \in \{ 1,0,-3,-1 \} \overset{1/4=2}{=\Longrightarrow}$$ $$\overset{1/4=2}{=\Longrightarrow} bc \in \{ 2,0,-6,-2 \}$$ Now we can count: $$\begin{array}{l} bc \overset{\text{Has 6 solutions}}{=======} {\color{white}-}2 & \Longrightarrow & a\overset{\text{Has 1 solution}}{=======}{\color{white}-}4 & \Longrightarrow & \text{Has ${\color{white}6}$6 Solutions} \\ bc \overset{\text{Has 13 solutions}}{=======} {\color{white}-}0 & \Longrightarrow & a\overset{\text{Has 2 solutions}}{=======}{\color{white}-}0,1 & \Longrightarrow & \text{Has 26 Solutions} \\ bc \overset{\text{Has 6 solutions}}{=======} -6 & \Longrightarrow & a\overset{\text{Has 2 solutions}}{=======}{\color{white}-}3,5 & \Longrightarrow & \text{Has 12 Solutions} \\ bc \overset{\text{Has 6 solutions}}{=======} -2 & \Longrightarrow & a\overset{\text{Has 2 solutions}}{=======}-1,2 & \Longrightarrow & \text{Has 12 Solutions} \\ \end{array}$$ So totally we have: $$6+26+12+12=56 \ \text{ Solutions }$$ Also you may don't calculate $a$ and just notice the relation between "Number of solution & Sign of $\Delta$" is hold in any field.

Answer 2

Thanks to the comment by @ancientmathematician , any matrix similar to the given matrix would have $0$ and $1$ as eigenvalues with corresponding eigenvectors similar upto a nonzero multiple of the column vectors (as they are cancelled in the inverse i.e. since $P^{-1}AP=M$). Thus, as the first and second columns of any matrix in $\mathbb{F}_p$ can be chosen in $(p^2-1)(p^2-p), p=7$ ways, therefore the desired number of eigenvectors are $\frac{(p^2-1)(p^2-p)}{(p-1)(p-1)}$ where $p=7$. Hence, the answer is $56$