Animated Solution for Mathematics - Matrices and Determinants: Comprehension Passage

Our journey begins by defining our playground. We have a set $T_p$ of $2\times 2$ matrices of the form $A=\left[\begin{array}{cc}a& b\\ c& a\end{array}\right]$, where every element is chosen from the set $\{0,1,2,\dots ,p-1\}$.

Since we have three independent variables—$a$, $b$, and $c$—and each has $p$ possible choices, the total number of matrices in our universe is:

This is our total sample space. The determinant of this matrix is given by:

This simple expression is the key that unlocks every door in this problem.

A matrix is symmetric if $b=c$. The determinant becomes $\det (A)=a^2-b^2$, which factors as $(a-b)(a+b)$. For this to be divisible by $p$, we require:

Because $p$ is a prime number, this forces either $a\equiv b(\mathrm{mod}p)$ or $a\equiv -b(\mathrm{mod}p)$. If $a=b$, we have $p$ choices. If $a=-b$, we have $p-1$ choices (excluding the case $a=b=0$ to avoid double counting), yielding $2p-1$ total symmetric matrices.

A matrix is skew-symmetric if $A^T=-A$. This forces the diagonal elements to be zero ($a=0$) and the off-diagonal elements to be negatives ($b=-c$). The determinant becomes $b^2$, which is divisible by $p$ only if $b=0$.

This leads us back to the zero matrix, which we have already counted in the symmetric case. Thus, the total count for matrices that are symmetric, skew-symmetric, or both, such that the determinant is divisible by $p$, remains $2p-1$.

We now consider the condition where the trace is not divisible by $p$ and the determinant is divisible by $p$. The trace is $2a$. Since $p$ is an odd prime, $2a ot\equiv 0 \pmod{p}$ implies $a eq 0$.

This gives us $p-1$ choices for $a$. For the determinant, we require $bc\equiv a^2(\mathrm{mod}p)$. Since $a eq 0$, $a^2$ is non-zero, which forces both $b$ and $c$ to be non-zero.

For any fixed $a$ and any non-zero $b$, there is a unique $c$ such that $c\equiv a^2{b}^{-1}(\mathrm{mod}p)$. Since there are $p-1$ choices for $b$, there are $p-1$ choices for $c$. Multiplying these, we get:

Finally, we find the number of matrices where the determinant is NOT divisible by $p$. We use complementary counting by subtracting the number of matrices where $\det (A)\equiv 0(\mathrm{mod}p)$ from the total $p^3$.

We split the zero-determinant case based on $a$: 1. If $a=0$, then $\det (A)=-bc\equiv 0(\mathrm{mod}p)$. This occurs if $b=0$ or $c=0$, resulting in $2p-1$ matrices. 2. If $a eq 0$, we found there are $(p-1{)}^2$ such matrices.

Summing these gives $(2p-1)+(p^2-2p+1)=p^2$. Subtracting this from the total, the number of matrices where the determinant is not divisible by $p$ is: