Let $P = \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{pmatrix}$ and $I$ be the identity matrix of order 3. If $Q = [q_{ij}]$ is a matrix such that $P^{50} - Q = I$, then $\frac{q_{31} + q_{32}}{q_{21}}$ equals
Step-by-Step Solution
Key Concept: The matrix P is a lower triangular matrix. Calculate P^n to find the pattern for the entries. Specifically, P = I + N where N is a nilpotent matrix. Use the binomial expansion P^50 = (I + N)^50 = I + 50N + (50*49/2)N^2. Then Q = P^50 - I = 50N + 1225N^2. Extract the required entries q_31, q_32, q_21.
Let $P = I + N$, where $N = \begin{pmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{pmatrix}$. Note $N^2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 16 & 0 & 0 \end{pmatrix}$ and $N^3 = 0$. Then $P^{50} = (I+N)^{50} = I + 50N + \frac{50 \times 49}{2} N^2 = I + 50N + 1225N^2$. Thus $Q = P^{50} - I = 50N + 1225N^2$. $Q = 50 \begin{pmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{pmatrix} + 1225 \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 16 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 200 & 0 & 0 \\ 800 + 19600 & 200 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 200 & 0 & 0 \\ 20400 & 200 & 0 \end{pmatrix}$. So $q_{31} = 20400, q_{32} = 200, q_{21} = 200$. The value is $(20400 + 200) / 200 = 20600 / 200 = 103$.
Correct Answer: B