Q19 / 50 · Manipal (MET) 2024 · Mathematics

Let $D=\left|\begin{array}{ccc}n & n^2 & n^3 \\ n^2 & n^3 & n^5 \\ 1 & 2 & 3\end{array}\right|$, then $\lim _{n \rightarrow \infty} \frac{M_{11}+C_{33}}{\left(M_{13}\right)^2}$ is equal to (where $M_{i j}$ is the minor and $C_{i j}$ is the cofactor of $i^{\text {th }}$ row and $j^{\text {th }}$ column's element)

  • Correct answer
    A
    0
  • B
    -1
  • C
    -2
  • D
    3

Solution

$\because D=\left|\begin{array}{ccc} n & n^2 & n^3 \\ n^2 & n^3 & n^5 \\ 1 & 2 & 3 \end{array}\right|$ Then, $M_{11}=\left|\begin{array}{cc}n^3 & n^5 \\ 2 & 3\end{array}\right|=3 n^3-2 n^5$ and $C_{33}=\left|\begin{array}{cc}n & n^2 \\ n^2 & n^3\end{array}\right|=n^4-n^4=0$ $\begin{aligned} M_{13} & =\left|\begin{array}{cc} n^2 & n^3 \\ 1 & 2 \end{array}\right|=2 n^2-n^3 \\ \therefore \lim _{n \rightarrow \infty} & =\frac{M_{11}+C_{33}}{\left(M_{13}\right)^2}=\lim _{n \rightarrow \infty} \frac{n^3\left(3-2 n^2\right)+0}{n^4(2-n)^2} \\ & =\lim _{n \rightarrow \infty} \frac{3-2 n^2}{n\left[4+n^2-4 n\right]}=\lim _{n \rightarrow \infty} \frac{n^2\left(\frac{3}{n^2}-2\right)}{n^3\left(\frac{4}{n^2}+1-\frac{4}{n}\right)} \end{aligned}$ put $n=\infty$ $=\frac{\left(\frac{3}{\infty^2}-2\right)}{\infty\left(\frac{4}{\infty^2}+1-\frac{4}{\infty}\right)}=\frac{-2}{1 \times \infty}=0$

Hence, the correct answer is option: 'A'.

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