Animated Solution for Mathematics - Matrices and Determinants: The number of matrices $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, where $a,b,c,d\in \{-1,0,1,2,3,\dots ,10\}$, such that $A=A^{-1}$, is

Imagine you are standing before a mirror. The reflection you see is a perfect, inverted copy of yourself. In the world of linear algebra, some matrices possess this exact property—they are their own inverse. These are called involutory matrices.

We are analyzing the matrix: where the elements are chosen from the set $S=\{-1,0,1,\dots ,10\}$.

The problem provides a beautiful starting point: $A=A^{-1}$. If we multiply both sides by $A$, we get $A\cdot A=A\cdot A^{-1}$, which simplifies to:

This is our North Star. It tells us that applying the transformation $A$ twice brings us back to the identity matrix:

Now, let us compute $A^2$ explicitly. By multiplying the matrix by itself, we obtain:

Equating this to the identity matrix gives us four powerful equations: 1) $a^2+bc=1$ 2) $b(a+d)=0$ 3) $c(a+d)=0$ 4) $bc+d^2=1$

Notice the symmetry! Equations (2) and (3) both feature the term $(a+d)$. This is our fork in the road.

Case I

We must split our analysis into two logical cases. Case I: $a+d=0$.

If $a+d=0$, then $d=-a$. Since $a$ and $d$ must be in our set $S=\{-1,0,1,\dots ,10\}$, we check the possibilities: If $a=1$, then $d=-1$. If $a=0$, then $d=0$. * If $a=-1$, then $d=1$.

For $a=1$ and $d=-1$, the equation $a^2+bc=1$ becomes $1+bc=1$, so $bc=0$. This means either $b=0$ or $c=0$. Counting these pairs, we have 12 choices for $b$ (with $c=0$) and 12 for $c$ (with $b=0$), minus the overlap where both are zero, giving us 23 pairs.

The same logic applies to $a=-1$ and $d=1$, yielding another 23 pairs. For $a=0$ and $d=0$, we get $bc=1$, which only happens if $(b,c)=(1,1)$ or $(-1,-1)$, giving us 2 more pairs.

Total for Case I: $23+23+2=48$.

Case II

Case II: $a+d eq 0$.

If $a+d eq 0$, then equations (2) and (3) force $b=0$ and $c=0$. Substituting these into our diagonal equations, we get $a^2=1$ and $d^2=1$.

This means $a$ and $d$ must be $1$ or $-1$. However, we must satisfy $a+d eq 0$. If $a=1$ and $d=-1$, the sum is zero, which is forbidden. Thus, $a$ and $d$ must have the same sign.

We have two possibilities: $(1,1)$ and $(-1,-1)$. This gives us 2 more matrices.

Adding our results together, we have $48+2=50$.

We have successfully navigated the constraints, avoided the double-counting traps, and arrived at the elegant solution of 50. Mathematics is not just about numbers; it is about finding the structure within the chaos.