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

Concept: 
• \(M_{ij}\) = minor 
• \(C_{ij} = (-1)^{i+j} M_{ij}\) 

Step 1: Compute minors \[ M_{11} = \begin{vmatrix} n^3 & n^5 \\ 2 & 3 \end{vmatrix} = 3n^3 - 2n^5 \sim -2n^5 \] \[ M_{13} = \begin{vmatrix} n^2 & n^3 \\ 1 & 2 \end{vmatrix} = 2n^2 - n^3 \sim -n^3 \] \[ C_{33} = (+1) M_{33}, \quad M_{33} = \begin{vmatrix} n & n^2 \\ n^2 & n^3 \end{vmatrix} = n \cdot n^3 - n^2 \cdot n^2 = 0 \] \[ \Rightarrow C_{33} = 0 \] 

Step 2: Substitute in expression \[ M_{11} + C_{33} \sim -2n^5 \] \[ (M_{13})^2 \sim (-n^3)^2 = n^6 \] 

Step 3: Limit \[ \lim_{n \to \infty} \frac{-2n^5}{n^6} = \lim_{n \to \infty} \frac{-2}{n} = 0 \]