Let $P$ be an odd prime number and $T_p$ be the following set of $2 × 2$ matrices, find following

Let $P$ be an odd prime number and $T_p$ be the following set of $2 × 2$ matrices :

$$T_p=\left\{\begin{bmatrix}a&b\\c&a\end{bmatrix};a,b,c \in \{0,1,2\cdots\cdots P-1\}\right\}$$

i) The number of $A$ in $T_p$ such that the trace of $A$ is not divisible by $P$ but $|A|$ is divisible by $P$ is

$a)$ $(P-1)(P^2-P+1)\quad$ $b)$ $P^3-(P-1)^2\quad$ $c)$ $(P-1)^2\quad$ $d)$ $(P-1)(P^2-2)$

ii) The number of $A$ in $T_p$ such that $|A|$ is not divisible by $P$ is

$a)$ $2P^2\quad$ $b)$ $P^3-5P\quad$ $c)$ $P^3-3P\quad$ $d)$ $P^3-P^2$

My attempt is as follows:-

$$n(T_p)=P^3$$

We can divide this set into types of matrices where $x$ is no of matrices with determinant divisible by $P$ and $y$ is no of matrices with determinant not divisible by $P$

$$x+y=P^3\tag{1}$$

Again matrices with determinant divisible by $P$ can be divided into two types of matrices where $u$ denotes no of matrices which have trace not divisible by $P$ and $v$ denotes no of matrices which have trace divisible by $P$

$$u+v=x\tag{2}$$

One can see that $u$ is the answer of first part and $y$ is the answer of second part

Using equation $1$

$$u+v=P^3-y$$ $$u+y=P^3-v\tag{3}$$

If we can calculate $v$, then by trial and error we can guess what $u$ and $y$ is from the given options

$v$ is no of matrices which have trace divisible by $P$ and determinant divisible by $P$

As we can see trace of the matrix $A$ is $T_{r}=2a$. For $T_{r}$ to be divisible by $P$, $2$ or $a$ has to divisible by $P$. Clearly $2$ cannot be divisible by $P$ as $P$ is odd prime number. So only choice is $a$, but $a<P$ so only for $a=0$, $T_r$ will be divisible by $P$

So under these conditions matrix A looks like $A=\begin{bmatrix}0&b\\c&0\end{bmatrix}$

$|A|=-bc$, $|A|$ will be divisible by $P$ under three cases:-

Case $1$: $b=0,c=0$

Case $2$: $b=0,c\in \{1,2\cdots\cdots P-1\}$

Case $3$: $c=0,b\in \{1,2,\cdots\cdots P-1\}$

So no of matrices which have trace divisible by $P$ and determinant divisible by $P$ are $2P-1$

$$v=2P-1$$

Putting the value of $v$ in equation $3)$

$$u+y=P^3-2P+1$$

Let's try with $u=(P-1)(P^2-P+1)$

$$y=P^3-2P+1-(P-1)(P^2-P+1)$$ $$y=P^3-2P+1-(P^3-P^2+P-P^2+P-1)$$ $$y=-2P+1+2P^2-2P+1$$ $$y=2P^2-4P+2$$ $$y=2(P^2+1-2P)$$ $$y=2(P-1)^2\quad \text { not given in options }$$

Let's try with $u=P^3-(P-1)^2$

$$y=P^3-2P+1-(P^3-P^2-1+2P)$$ $$y=-2P+1+P^2-2P+1$$ $$y=P^2-4P+2\quad \text { not given in options }$$

Let's try with $u=(P-1)^2$

$$y=P^3-2P+1-(P^2+1-2P)$$ $$y=P^3-P^2\quad \text { given as $d$ option } $$

This can be the answer

Let's try with last option $u=(P-1)(P^2-2)$

$$y=P^3-2P+1-(P^3-2P-P^2+2)$$ $$y=P^2-1\quad \text { not given in options }$$

So answer of first part is $(P-1)^2$ and answer of second part is $P^3-P^2$

So this is the totally objective approach. If I had not been given any options, I would not have been able to solve this. Is there any actual way to solve this problem, please share your ideas. I was stuck in this problem for a long time thinking of somehow to calculate but at last I came up with this approach.

Solution 1:

Here is how one can solve this type of problem.

Part (i)

Working modulo $p$, you require $a\ne 0$ and $a^2=bc$.

There are $p-1$ choices for $a$. Then $bc\ne0$ and so there are $p-1$ possible choices for $b$. For each choice, by the uniqueness of inverse, there is just one choice for $c$.

The total number of choices is $(p-1)^2$.

Part (ii)

The number with Tr$(a)\ne 0$ is $(p-1)p^2$.($a$ must be non-zero and $b,c$ can be any of the elements.)

So the number with Tr$(a)\ne 0$ and det non-zero is just $(p-1)p^2$ minus $(p-1)^2$ (The previous answer to part (i)).

We must add on the number of matrices with Tr$(a)= 0$ and det non-zero, which is $(p-1)^2$.

The total number of choices is $(p-1)p^2$.