Animated Solution for Mathematics - Matrices and Determinants: Let the matrix $A=\left[\begin{array}{ccc}1& 0& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]$ satisfy $A^n=A^{n-2}+A^2-I$ for $n\ge 3$. Then the sum of all the elements of $A^{50}$ is :-

Unlocking the Power of Fifty

Imagine you are standing before a massive, intimidating matrix:

The question asks for the sum of all elements in $A^{50}$. Your first instinct might be to panic, but look at the recurrence relation: $A^n=A^{n-2}+A^2-I$.

This is not just an equation; it is a map. It tells us that the powers of $A$ are not growing chaotically, but in a highly structured, predictable way.

The Foundation

Before we can conquer the fiftieth power, we must understand the base. The recurrence relation depends on $A^2$, so let us calculate that first.

Multiplying $A$ by itself, we get:

This matrix $A^2$ is our building block. Keep it safe; we will need it later.

The Telescoping Magic

Now, look at the recurrence relation again: $A^n-A^{n-2}=A^2-I$. This is the 'Aha!' moment.

The difference between any two powers of $A$ that are two steps apart is always the same constant matrix, $D=A^2-I$. This is the definition of a telescoping series.

If we write this out for even powers from $n=4$ to $n=50$, we get:

When we add all these equations together, the magic happens. The $A^4$ cancels with $-A^4$, the $A^6$ cancels with $-A^6$, and this continues until only the first negative term ($-A^2$) and the last positive term ($A^{50}$) remain.

We are left with:

Since there are 24 terms in this sum, we get:

The Final Calculation

Now, it is just a matter of simple algebra. Isolate $A^{50}$:

We know $A^2$ and we know $I$. Let us plug them in:

Performing the scalar multiplication and subtraction:

Finally, the sum of all elements is $1+0+0+25+1+0+25+0+1=53$.

There you have it! By recognizing the pattern and using the telescoping sum, we turned a seemingly impossible problem into a simple, elegant solution. Never let the size of the exponent intimidate you; look for the structure, and the math will reveal itself.