Animated Solution for Mathematics - Matrices and Determinants: For the matrices $A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$ and $B=\left[\begin{array}{cc}-29& 49\\ -13& 18\end{array}\right]$, if $(A^{15}+B)\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$, then among the following which one is true?

Welcome, future engineer! Today, we are going to tackle a problem that looks intimidating at first glance but hides a beautiful, elegant secret.

We are given two matrices, $A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$ and $B=\left[\begin{array}{cc}-29& 49\\ -13& 18\end{array}\right]$, and we are asked to solve:

The moment you see $A^{15}$, I want you to take a deep breath. Do not reach for your pen to multiply matrix $A$ fifteen times; that is the trap. In the world of JEE Advanced, whenever you see a high power of a matrix, it is a signal to pause and look for a pattern.

Our first mission is to find the soul of matrix $A$. We start by calculating $A^2=A\cdot A$:

Now, let us find $A^3=A^2\cdot A$:

Now, look at the sequence of matrices: $A^1=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$, $A^2=\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right]$, and $A^3=\left[\begin{array}{cc}7& -12\\ 3& -5\end{array}\right]$.

Do you see it? The bottom-left element follows $1,2,\mathrm{3...}$ which is simply $n$. The top-right element follows $-4,-8,-\mathrm{12...}$ which is $-4n$.

The top-left element follows $3,5,\mathrm{7...}$ which is $2n+1$. The bottom-right element follows $-1,-3,-\mathrm{5...}$ which is $-2n+1$.

We have cracked the code! The general form is:

With our master key $A^n$ in hand, finding $A^{15}$ is trivial. We substitute $n=15$:

Now, we add matrix $B$ to this result:

This result is stunningly symmetric. When we multiply this by the column vector $\left[\begin{array}{c}x\\ y\end{array}\right]$, we get the system:

This simplifies to $2x=11y$, or $x=\frac{11y}{2}$.

Testing our options, we find that $x=11$ and $y=2$ satisfies this perfectly. You see? By looking for the pattern, we turned a nightmare calculation into a simple, logical flow. Keep this mindset, and you will conquer any problem the exam throws at you!