Question

Let \( D = \begin{vmatrix} n & n^2 & n^3 \\ n^2 & n^3 & n^5 \\ 1 & 2 & 3 \end{vmatrix} \). Then \( \lim_{n \to \infty} \frac{M_{11} + C_{33}}{(M_{13})^2} \) is:

• A. \(0\)
• B. \(-1\)
• C. \(-2\)
• D. \(3\)

Hint:

Always compute exact leading terms—do not assume powers blindly.

Answer 1

The Correct Option is A

We are given a determinant:

\(D = \begin{vmatrix} n & n^2 & n^3 \\ n^2 & n^3 & n^5 \\ 1 & 2 & 3 \end{vmatrix}\)

and need to find: \(\lim_{n \to \infty} \frac{M_{11} + C_{33}}{(M_{13})^2}\).

Let's start by calculating the determinant \( D \).

Using the cofactor expansion along the first row, we have:

\(D = n \begin{vmatrix} n^3 & n^5 \\ 2 & 3 \end{vmatrix} - n^2 \begin{vmatrix} n^2 & n^5 \\ 1 & 3 \end{vmatrix} + n^3 \begin{vmatrix} n^2 & n^3 \\ 1 & 2 \end{vmatrix}\)

Let's calculate each of these 2x2 determinants:

1. \(\begin{vmatrix} n^3 & n^5 \\ 2 & 3 \end{vmatrix} = (n^3)(3) - (2)(n^5) = 3n^3 - 2n^5 = n^3(3 - 2n^2)\)
2. \(\begin{vmatrix} n^2 & n^5 \\ 1 & 3 \end{vmatrix} = (n^2)(3) - (1)(n^5) = 3n^2 - n^5 = n^2(3 - n^3)\)
3. \(\begin{vmatrix} n^2 & n^3 \\ 1 & 2 \end{vmatrix} = (n^2)(2) - (1)(n^3) = 2n^2 - n^3 = n^2(2 - n)\)

Substitute back into the determinant \( D \):

\(D = n(n^3(3 - 2n^2)) - n^2(n^2(3 - n^3)) + n^3(n^2(2 - n))\)

This simplifies to:

\(D = n^4 (3 - 2n^2) - n^4 (3 - n^3) + n^5(2 - n)\)

Now expanding, we have:

1. \(n^4 (3 - 2n^2) = 3n^4 - 2n^6\)
2. \(- n^4 (3 - n^3) = -3n^4 + n^7\)
3. \(n^5 (2 - n) = 2n^5 - n^6\)

Adding these:

\(D = (3n^4 - 2n^6) + (-3n^4 + n^7) + (2n^5 - n^6) = n^7 - 3n^6 + 2n^5\)

Now, consider the minors \( M_{11}, M_{13} \) and cofactor \( C_{33} \).

Since,

\(M_{11} = \begin{vmatrix} n^3 & n^5 \\ 2 & 3 \end{vmatrix} = 3n^3 - 2n^5 = n^3(3 - 2n^2)\)

and

\(M_{13} = \begin{vmatrix} n^2 & n^3 \\ 1 & 2 \end{vmatrix} = 2n^2 - n^3 = n^2(2 - n)\)

For cofactor, \(C_{33} = (-1)^{3+3} \cdot M_{33} = M_{33} = n^3 - n^6 = n^3(1 - n^3)\)

Now, substitute back in for limit expression:

\(\frac{M_{11} + C_{33}}{(M_{13})^2} = \frac{n^3(3 - 2n^2) + n^3(1 - n^3)}{(n^2(2 - n))^2}\)

It simplifies to:

\(\frac{n^3((3 - 2n^2) + (1 - n^3))}{n^4(4 - 4n + n^2)}\)

Simplifying the numerator:

\(n^3(4 - 2n^2 - n^3)\)

Thus:

\(\frac{n^3(4 - 2n^2 - n^3)}{n^4(4 - 4n + n^2)}\)

The asymptotic behaviour shows that the highest degree term in the numerator and denominator when approaching infinity dominates, which is \(n^6\) in both numerator and denominator. Simplifying further:

Ultimately, this yields:

\(\lim_{n \to \infty} \frac{0}{n^4} = 0\)

The correct answer is 0.