JEE Advanced 2018 Paper 1 Mathematics: Paragraph 1

Since $M$ has positive entries, $M^n(1,0)^T$ aligns with the dominant eigenvector. The characteristic equation is $\lambda^2-3\lambda+1=0$, so the dominant eigenvalue is $\lambda=\frac{3+\sqrt5}2$. For an eigenvector $(x,y)$: $2x+y=\lambda x$, so $y=(\lambda-2)x=\frac{\sqrt5-1}2x$. Therefore $\frac xy=\frac2{\sqrt5-1}=\frac{\sqrt5+1}2$. Numerically $x_n/y_n=5/3,\,13/8,\,34/21,\dots$ approaches $1.618$, consistent with this. Option B is the eigenvalue itself (the ratio $x_{n+1}/x_n$, not $x_n/y_n$), option C is the reciprocal $y_n/x_n$ and option D is only a rough guess.

Let $v_n=M^n e_1$ with $e_1=(1,0)^T$. Then $\sum_{n\ge0}\frac{v_n}{3^n}=\left(I-\frac M3\right)^{-1}e_1$, which converges since the dominant eigenvalue $\frac{3+\sqrt5}2\approx2.618<3$. Now $I-\frac M3=\frac13\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$ with determinant $\frac19(2-1)=\frac19$, so its inverse is $3\begin{pmatrix}2&1\\1&1\end{pmatrix}$. Applying to $e_1$ gives $(6,3)^T$. Thus $\sum x_n/3^n=6$ and $\sum y_n/3^n=3$, and the required sum is $9$.