Problem 43 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(A=\left[\begin{array}{cc}1 & \frac{\alpha}{n} \\ -\frac{\alpha}{n} & 1\end{array}\right]\), then (A) \(\lim _{n \rightarrow \infty} A^{n}=0\) (B) \(\lim _{n \rightarrow \infty} \frac{1}{n} A^{n}=0\) (C) \(\lim _{n \rightarrow \infty} \frac{1}{n^{2}} A^{n}=0\) (D) None of these

Step-by-Step Solution

Answer

Option (C) is correct: \( \lim_{n \to \infty} \frac{1}{n^2} A^n = 0 \).

Step 1: Calculate the Determinant of A

To understand the characteristics of matrix \( A \), first calculate its determinant. The determinant \( \text{det}(A) \) for a matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( ad - bc \). Here, \( a = 1 \), \( b = \frac{\alpha}{n} \), \( c = -\frac{\alpha}{n} \), and \( d = 1 \). Thus, \( \text{det}(A) = (1)(1) - \left(\frac{\alpha}{n}\right)\left(-\frac{\alpha}{n}\right) = 1 + \frac{\alpha^2}{n^2} \).

Step 2: Analyze the Eigenvalues

To explore the matrix powers, compute the eigenvalues of \( A \). The eigenvalues are the roots of the characteristic equation \( \lambda^2 - (a+d)\lambda + (ad-bc) = 0 \). Here, \( a+d = 2 \) and \( ad-bc = 1 + \frac{\alpha^2}{n^2} \). Solving \( \lambda^2 - 2\lambda + (1 + \frac{\alpha^2}{n^2}) = 0 \) yields the eigenvalues: \( \lambda = 1 \pm \frac{\alpha}{n}i \).

Step 3: Evaluate Matrix Powers

For large \( n \), observe the behavior of \( A^n \). Since the eigenvalues are \( 1 \pm \frac{\alpha}{n}i \), the magnitude \( |\lambda| = 1 \). As \( n \to \infty \), these eigenvalues do not converge to zero, suggesting the matrix \( A^n \) will not converge to zero.

Step 4: Validate Each Option

Examine each option: (A) \( \lim_{n \to \infty} A^n = 0 \) is false because \( A^n \) does not approach zero. (B) \( \lim_{n \to \infty} \frac{1}{n} A^n = 0 \) appears false because \( \frac{1}{n} A^n \) remains nonzero for large \( n \). (C) \( \lim_{n \to \infty} \frac{1}{n^2} A^n = 0 \) is plausible since dividing by \( n^2 \) neutralizes the \( n \) factor. Calculate: For \( \lambda = 1 \pm \frac{\alpha}{n}i \), \( |\lambda| = 1 \). Thus, dividing \( A^n \) by \( n^2 \) makes each entry approach zero as \( n \to \infty \).

Step 5: Conclusion

By validating each option, (C) is the correct answer as \( \lim_{n \to \infty} \frac{1}{n^2} A^n = 0 \) by scaling the rapidly oscillating terms effectively to zero.

Key Concepts

Determinant Calculation

To engage with any matrix, it's essential to calculate its determinant as the initial step. This involves understanding the characteristics of the matrix, which can provide insight into its properties. For a basic 2x2 matrix \[\begin{bmatrix}a & b \c & d\end{bmatrix}\]the determinant is calculated using the formula: \[\text{det}(A) = ad - bc\].For our specific matrix \[A = \begin{bmatrix} 1 & \frac{\alpha}{n} \ -\frac{\alpha}{n} & 1 \end{bmatrix}\], where each element relates either directly to 1 or is scaled by \(\frac{\alpha}{n}\), the determinant becomes \[1 - \left(-\frac{\alpha^2}{n^2}\right) = 1 + \frac{\alpha^2}{n^2}\].This calculation reveals that the determinant remains positive as long as \(\alpha\) is real, providing a preliminary understanding that the matrix is invertible and thus has significant mathematical structure. By retaining the determinant greater than zero, the matrix resists collapsing, hinting at stable transformations.

Eigenvalues

In the realm of matrices, eigenvalues play a crucial role, especially when exploring matrix powers and limits. Eigenvalues are special numbers associated with a matrix that emerge from the characteristic equation derived from the matrix itself. For a 2x2 matrix like \[A = \begin{bmatrix} 1 & \frac{\alpha}{n} \ -\frac{\alpha}{n} & 1 \end{bmatrix}\], the process involves identifying roots of the characteristic polynomial: \[\lambda^2 - (a+d)\lambda + (ad-bc) = 0\]. Here, where \(a + d = 2\) and \(ad - bc = 1 + \frac{\alpha^2}{n^2}\), solving this equation yields eigenvalues: \[\lambda = 1 \pm \frac{\alpha}{n}i\].Understanding that these eigenvalues have magnitudes of 1, due to the formula for magnitude being \(|\lambda| = \sqrt{\text{Real}^2 + \text{Imaginary}^2}\), it leads to insight regarding matrix behavior under repetitive separation or application. Particularly, since the magnitudes do not diminish, this indicates persistence rather than decay in the repetitive matrix applications such as matrix powers.

Matrix Powers

Matrix powers provide a means to understand how a matrix behaves when multiplied by itself many times, particularly relevant in finding matrix limits as \(n\) approaches infinity. This incorporates the behavior of eigenvalues, given the exponential nature of such powers. A notable aspect is whether the matrix power will converge to zero, oscillate, or stabilize. For our matrix \[A = \begin{bmatrix} 1 & \frac{\alpha}{n} \ -\frac{\alpha}{n} & 1 \end{bmatrix}\], we have eigenvalues \(\lambda = 1 \pm \frac{\alpha}{n}i\) with \(|\lambda| = 1\). Despite oscillatory properties indicated by these eigenvalues, multiplication of such a matrix by itself doesn't lead to convergence to 0.

• This behavior is studied by looking at sequences like \(A^n\) versus factorized sequences like \(\frac{1}{n}A^n\) or \(\frac{1}{n^2}A^n\).
• The latter, \(\frac{1}{n^2}A^n\), suggests that with significant scaling (i.e., division by \(n^2\)), the rapid oscillations diminish, eventually leading the matrix elements towards zero.

This insight is used to evaluate limits rigorously, explaining why in this exercise, the correct conclusion, \(\lim_{n \to \infty} \frac{1}{n^2} A^n = 0\), holds true.