Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{cc}2& -2\\ 1& -1\end{array}\right]$ and $B=\left[\begin{array}{cc}-1& 2\\ -1& 2\end{array}\right]$. Then the number of elements in the set $\{(n,m):n,m\in \{1,2,\dots ,10\}\text{ and }n{A}^n+m{B}^m=I\}$ is

Unmasking the Idempotent Matrix

Imagine you are sitting in the examination hall, the clock is ticking, and you see a problem involving $A^n$ and $B^m$ where $n$ and $m$ go up to ten. Your first instinct might be panic.

You might think, "Do I need to calculate $A^{10}$? That will take forever!" But here is the secret of JEE Advanced: complexity is often a mask.

The problem setters are not testing your ability to perform tedious arithmetic; they are testing your ability to spot the elegant, underlying structure. Let us peel back the layers of this problem together.

The Discovery of Idempotence

When you encounter a matrix raised to a variable power, your first move should always be to test the waters. Calculate $A^2$.

Let us perform the multiplication:

As we multiply the first row by the first column, we get $(2)(2)+(-2)(1)=4-2=2$. For the first row, second column, we get $(2)(-2)+(-2)(-1)=-4+2=-2$.

Continuing this, we find that the resulting matrix is exactly $\left[\begin{array}{cc}2& -2\\ 1& -1\end{array}\right]$. We have just discovered that $A^2=A$.

In the world of linear algebra, a matrix that satisfies $A^2=A$ is called an idempotent matrix. It is a matrix that, when multiplied by itself, remains unchanged.

This means $A^3=A^2\cdot A=A\cdot A=A$. By induction, $A^n=A$ for all $n\ge 1$. The monster has been tamed.

We apply the same logic to matrix $B=\left[\begin{array}{cc}-1& 2\\ -1& 2\end{array}\right]$. Calculating $B^2$, we find:

Just like $A$, matrix $B$ is also idempotent. This means $B^m=B$ for all $m\ge 1$.

Collapsing the Equation

Now, look at our original equation: $n{A}^n+m{B}^m=I$. Because we have proven that $A^n=A$ and $B^m=B$, the equation collapses into a simple linear form: $nA+mB=I$.

This is the beauty of mathematics—taking a seemingly impossible problem and reducing it to a simple linear system. Let us substitute the matrices:

Distributing the scalars $n$ and $m$, we get:

Combining these, we arrive at the system:

The Final Resolution

We now have four equations, but we only need to solve for two variables. Let us look at the bottom-left element: $n-m=0$, which implies $n=m$.

This is a massive breakthrough. Now, substitute $n=m$ into the top-left element equation: $2n-m=1$.

Since $m=n$, this becomes $2n-n=1$, which simplifies to $n=1$. Consequently, $m=1$.

We have found our candidate pair $(1,1)$. But wait! A true mathematician never stops at the first result. We must verify this against the other elements.

Check the top-right element: $-2n+2m=-2(1)+2(1)=0$. This matches the identity matrix! Check the bottom-right element: $-n+2m=-1+2(1)=1$. This also matches!

The pair $(1,1)$ is the unique solution. Since $n$ and $m$ are restricted to the set $\{1,2,\dots ,10\}$, and our solution $(1,1)$ fits perfectly, we have found the only valid pair.

The number of elements in the set is exactly 1. You see? The complexity was just a test of your patience and your ability to look for patterns. Keep this mindset, and you will conquer any problem the JEE throws at you.