Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left(\begin{array}{cc}x& 1\\ 1& 0\end{array}\right)$, $x\in R$ and $A^4=[a_{ij}]$. If $a_{11}=109$, then $a_{22}$ is equal to

Welcome, fellow traveler on the road to JEE Advanced. Today, we are not just solving a matrix problem; we are peeling back the layers of a mathematical onion.

When you see a matrix $A=\left(\begin{array}{cc}x& 1\\ 1& 0\end{array}\right)$ raised to the fourth power, your first instinct might be to panic. But I want you to take a deep breath. In the world of linear algebra, complexity is often just a mask for underlying simplicity.

We begin by calculating $A^2$. It is the foundation of our tower. By performing the multiplication $A\cdot A=\left(\begin{array}{cc}x& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}x& 1\\ 1& 0\end{array}\right)$, we obtain:

Look at that structure. It is elegant, isn't it? The symmetry is already starting to emerge. We are not just crunching numbers; we are observing the evolution of a system.

Now, we move to $A^3$. We take our result from $A^2$ and multiply it by $A$ once more:

Do you see the pattern? The elements are shifting, growing, and rearranging themselves like dancers in a choreographed routine. We are now one step away from our target, $A^4$.

Finally, we compute $A^4$ by multiplying $A^3$ by $A$:

We have arrived. The matrix $A^4$ is fully revealed. The problem tells us that $a_{11}=109$. Looking at our matrix, we see that $a_{11}=x^4+3x^2+1$.

Thus, we set up our equation:

This is where the magic happens. By subtracting $109$ from both sides, we get $x^4+3x^2-108=0$.

Let us introduce a substitution to make this look friendly. Let $t=x^2$. Our equation transforms into a simple quadratic:

Factoring this quadratic is a joy. We look for two numbers that multiply to $-108$ and add to $3$. Those numbers are $12$ and $-9$.

So, we have $(t+12)(t-9)=0$. This gives us two potential values for $t$: $t=-12$ or $t=9$.

Recall that $t=x^2$. Since $x$ is a real number, $x^2$ cannot be negative. We must discard $t=-12$.

We are left with $x^2=9$. Now, look back at our matrix $A^4$. The element $a_{22}$ is simply $x^2+1$. Substituting our value of $x^2$, we get:

And there it is. The complexity dissolves, leaving behind a clean, satisfying integer. You didn't just solve a problem; you navigated a logical path through a forest of variables. The final answer is 10.