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(A) The matrix $A$ is a diagonal matrix where the diagonal elements are $a_{ii} = k^i$ for $i = 1, 2, \dots, n$.The trace of $A$, denoted by $m$, is t

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$18$

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(A) The matrix $A$ is a diagonal matrix where the diagonal elements are $a_{ii} = k^i$ for $i = 1, 2, \dots, n$.The trace of $A$, denoted by $m$, is the sum of its diagonal elements:$m = \sum_{i=1}^{n} k^i = k + k^2 + \dots + k^n = \frac{k(1-k^n)}{1-k}$.We are given the limit: $\lim_{k \rightarrow 1} \frac{n-m}{1-k} = 171$.Substituting $m$: $\lim_{k \rightarrow 1} \frac{n - \frac{k(1-k^n)}{1-k}}{1-k} = \lim_{k \rightarrow 1} \frac{n(1-k) - (k - k^{n+1})}{(1-k)^2} = 171$.Using $L$'Hospital's rule (differentiating numerator and denominator with respect to $k$):Numerator derivative: $\frac{d}{dk} [n - nk - k + k^{n+1}] = -n - 1 + (n+1)k^n$.Denominator derivative: $\frac{d}{dk} [(1-k)^2] = 2(1-k)(-1) = -2(1-k)$.Applying $L$'Hospital's rule again:$\lim_{k \rightarrow 1} \frac{-n - 1 + (n+1)k^n}{-2(1-k)} = \lim_{k \rightarrow 1} \frac{(n+1)n k^{n-1}}{2} = 171$.$\frac{n(n+1)}{2} = 171 \Rightarrow n^2 + n - 342 = 0$.Solving the quadratic equation: $(n+19)(n-18) = 0$.Since $n > 0$, we have $n = 18$.

Let $A = (a_{ij})$ be an $n \times n$ matrix defined by $a_{ij} = \begin{cases} k^i, & \forall i=j \\ 0, & \text{otherwise} \end{cases}$. If $m = \text{trace of } A$ and $\lim_{k \rightarrow 1} \frac{n-m}{1-k} = 171$, then the value of $n$ is:

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