Question ID: #695
For the matrices $A=\begin{bmatrix}3&-4\\ 1&-1\end{bmatrix}$ and $B=\begin{bmatrix}-29&49\\ -13&18\end{bmatrix},$ if $(A^{15}+B)\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}$, then among the following which one is true?
- (1) $x=5, \quad y=7$
- (2) $x=18, \quad y=11$
- (3) $x=11, \quad y=2$
- (4) $x=16, \quad y=3$
Solution:
Calculate the powers of matrix $A$:
$$A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$$
$$A^2 = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 9-4 & -12+4 \\ 3-1 & -4+1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$$
Observing the pattern, we can generalize for $A^n$:
$$A^n = \begin{bmatrix} 2n+1 & -4n \\ n & -2n+1 \end{bmatrix}$$
For $n=15$:
$$A^{15} = \begin{bmatrix} 2(15)+1 & -4(15) \\ 15 & -2(15)+1 \end{bmatrix} = \begin{bmatrix} 31 & -60 \\ 15 & -29 \end{bmatrix}$$
Now compute $A^{15} + B$:
$$A^{15} + B = \begin{bmatrix} 31 & -60 \\ 15 & -29 \end{bmatrix} + \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix}$$
$$A^{15} + B = \begin{bmatrix} 31-29 & -60+49 \\ 15-13 & -29+18 \end{bmatrix} = \begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix}$$
Given equation $(A^{15}+B)\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}$:
$$\begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
This yields the linear equation:
$$2x – 11y = 0 \Rightarrow 2x = 11y \Rightarrow \frac{x}{y} = \frac{11}{2}$$
Checking the options:
(1) $5/7 \ne 11/2$
(2) $18/11 \ne 11/2$
(3) $11/2 = 11/2$ (Satisfies the condition)
(4) $16/3 \ne 11/2$
Ans. (3)
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