Evaluate Expression Involving Elements of Matrix Difference

To find the value of the given expression, we first express the matrix $P$ as the sum of the identity matrix $I$ and a strictly lower triangular matrix $A$:

$P = I + A$

where

$$I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \quad \text{and} \quad A = \begin{bmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{bmatrix}$$

Now, let us calculate the higher powers of matrix $A$:

$$A^2 = \begin{bmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 16 & 0 & 0 \end{bmatrix}$$ $$A^3 = A^2 \cdot A = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 16 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$$

Since $A^k = O$ for all $k \ge 3$, matrix $A$ is nilpotent of index $3$.

Since $I$ and $A$ commute ($IA = AI$), we can expand $P^{n} = (I + A)^n$ using the Binomial Theorem:

$P^n = I + nA + \frac{n(n-1)}{2!} A^2$

For $n = 50$:

$P^{50} = I + 50A + \frac{50 \times 49}{2} A^2 = I + 50A + 1225 A^2$

We are given that $P^{50} - Q = I$, which gives:

$Q = P^{50} - I = 50A + 1225 A^2$

Substituting the matrices $A$ and $A^2$ into the equation for $Q$:

$$Q = 50 \begin{bmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{bmatrix} + 1225 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 16 & 0 & 0 \end{bmatrix}$$ $$Q = \begin{bmatrix} 0 & 0 & 0 \\ 200 & 0 & 0 \\ 800 & 200 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 19600 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 200 & 0 & 0 \\ 20400 & 200 & 0 \end{bmatrix}$$

From matrix $Q = [q_{ij}]$, we identify the required elements:

• $q_{21} = 200$
• $q_{31} = 20400$
• $q_{32} = 200$

Thus, the ratio $\frac{q_{31} + q_{32}}{q_{21}}$ is:

$\frac{q_{31} + q_{32}}{q_{21}} = \frac{20400 + 200}{200} = \frac{20600}{200} = 103$

Hence, the correct option is (B).