For the matrices A=beginbmatrix3 & -4 \\ 1 & -1endbmatrix$A=\begin{bmatrix}3 & -4 \\ 1 & -1\end{bmatrix}$ and B=beginbmatrix-29 & 49 \\ -13 & 18endbmatrix$B=\begin{bmatrix}-29 & 49 \\ -13 & 18\end{bmatrix}$, if (A^15+B)beginbmatrixx\\ yendbmatrix=beginbmatrix0\\ 0endbmatrix$(A^{15}+B)\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}$, then among the following which one is true?

### Related Formula A = I + N implies A^n = (I+N)^n = I + nN$$A = I + N \implies A^n = (I+N)^n = I + nN$$ (provided N^2 = 0$N^2 = 0$, meaning N$N$ is nilpotent of order 2). ### Core Logic Evaluate powers of matrix A$A$ to detect a pattern. A = beginbmatrix3 & -4 \\ 1 & -1endbmatrix = beginbmatrix1 & 0 \\ 0 & 1endbmatrix + beginbmatrix2 & -4 \\ 1 & -2endbmatrix$$A = \begin{bmatrix}3 & -4 \\ 1 & -1\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} + \begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}$$ Let N = beginbmatrix2 & -4 \\ 1 & -2endbmatrix$N = \begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}$. Check N^2$N^2$: N^2 = beginbmatrix2 & -4 \\ 1 & -2endbmatrixbeginbmatrix2 & -4 \\ 1 & -2endbmatrix = beginbmatrix4-4 & -8+8 \\ 2-2 & -4+4endbmatrix = beginbmatrix0 & 0 \\ 0 & 0endbmatrix$$N^2 = \begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}\begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix} = \begin{bmatrix}4-4 & -8+8 \\ 2-2 & -4+4\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$$ ### Step 1: Compute A^15 Since N^2 = 0$N^2 = 0$, A^n = I + nN$A^n = I + nN$. A^15 = I + 15N = beginbmatrix1 & 0 \\ 0 & 1endbmatrix + 15beginbmatrix2 & -4 \\ 1 & -2endbmatrix = beginbmatrix1 & 0 \\ 0 & 1endbmatrix + beginbmatrix30 & -60 \\ 15 & -30endbmatrix$$A^{15} = I + 15N = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} + 15\begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} + \begin{bmatrix}30 & -60 \\ 15 & -30\end{bmatrix}$$ A^15 = beginbmatrix31 & -60 \\ 15 & -29endbmatrix$$A^{15} = \begin{bmatrix}31 & -60 \\ 15 & -29\end{bmatrix}$$ ### Step 2: Evaluate Matrix Operation A^15 + B = beginbmatrix31 & -60 \\ 15 & -29endbmatrix + beginbmatrix-29 & 49 \\ -13 & 18endbmatrix = beginbmatrix2 & -11 \\ 2 & -11endbmatrix$$A^{15} + B = \begin{bmatrix}31 & -60 \\ 15 & -29\end{bmatrix} + \begin{bmatrix}-29 & 49 \\ -13 & 18\end{bmatrix} = \begin{bmatrix}2 & -11 \\ 2 & -11\end{bmatrix}$$ Given (A^15 + B)beginbmatrixx \\ yendbmatrix = beginbmatrix0 \\ 0endbmatrix$(A^{15} + B)\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$: beginbmatrix2 & -11 \\ 2 & -11endbmatrixbeginbmatrixx \\ yendbmatrix = beginbmatrix0 \\ 0endbmatrix$$\begin{bmatrix}2 & -11 \\ 2 & -11\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$$ This yields a single independent equation: 2x - 11y = 0 implies 2x = 11y$$2x - 11y = 0 \implies 2x = 11y$$ Checking the options, for x=11, y=2$x=11, y=2$, 2(11) = 11(2) implies 22 = 22$2(11) = 11(2) \implies 22 = 22$. Thus, Option 3 satisfies the condition. ### Pattern Recognition Whenever you need a high power of a 2times 2$2\times 2$ matrix, check its trace and determinant. If |A-lambda I|=0$|A-\lambda I|=0$ yields lambda=1,1$\lambda=1,1$, then A$A$ can be written as I+N$I+N$ where N$N$ is nilpotent. This bypasses lengthy matrix multiplication. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants