$$\begin{gathered} \text{For matrices }A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]\text{and}\mathrm{ B}=\left[\begin{array}{cc}-29& 49\\ -13& 18\end{array}\right]\text{, if}(A^{15}+B)\cdot \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\text{, then among the following which one is correct} \\ \end{gathered}$$

The correct option is x = 11, y = 2.

Step 1: Determine the characteristic equation of matrix A
Given matrix:

A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]

The trace of matrix A is:

\text{tr}\left(A\right)=3+(-1)=2

The determinant of matrix A is:

\text{det}\left(A\right)=\left(3\right)(-1)-(-4)\left(1\right)=-3+4=1

The characteristic equation of matrix A is given by:

{\lambda }^2-\text{tr}\left(A\right)\lambda +\text{det}\left(A\right)=0\Rightarrow {\lambda }^2-2\lambda +1=0

According to the Cayley-Hamilton Theorem, every matrix satisfies its own characteristic equation:

A^2-2A+I=0\Rightarrow A^2=2A-I

Step 2: Find a general expression for Aⁿ
Let's find higher powers of A:

A^3=A\cdot A^2=A(2A-I)=2A^2-A=2(2A-I)-A=3A-2I

A^4=A\cdot A^3=A(3A-2I)=3A^2-2A=3(2A-I)-2A=4A-3I

By induction, for any positive integer n:

A^n=nA-(n-1)I

Substituting n = 15:

A^{15}=15A-14I

Now, calculate matrix A¹⁵:

A^{15}=15\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]-14\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]

A^{15}=\left[\begin{array}{cc}45& -60\\ 15& -15\end{array}\right]-\left[\begin{array}{cc}14& 0\\ 0& 14\end{array}\right]=\left[\begin{array}{cc}31& -60\\ 15& -29\end{array}\right]

Step 3: Calculate (A¹⁵ + B)
Adding matrix B to matrix A¹⁵:

A^{15}+B=\left[\begin{array}{cc}31& -60\\ 15& -29\end{array}\right]+\left[\begin{array}{cc}-29& 49\\ -13& 18\end{array}\right]

A^{15}+B=\left[\begin{array}{cc}31-29& -60+49\\ 15-13& -29+18\end{array}\right]=\left[\begin{array}{cc}2& -11\\ 2& -11\end{array}\right]

Step 4: Solve the given matrix equation
We are given:

(A^{15}+B)\cdot \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]

Substituting (A¹⁵ + B):

\left[\begin{array}{cc}2& -11\\ 2& -11\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]

Performing matrix multiplication gives the equation:

2x-11y=0\Rightarrow 2x=11y

Checking the values from the options, for x = 11 and y = 2:

2\left(11\right)-11\left(2\right)=22-22=0

Thus, x = 11 and y = 2 satisfy the equation correctly.