Question

Find the largest eigenvalue of the matrix $\begin{bmatrix} 5 & 4 \\1 & 2 \end{bmatrix}$

• A. 1
• B. 6
• C. 2
• D. 3

Hint:

To find eigenvalues, solve \( \det(A - \lambda I) = 0 \). For a 2x2 matrix, this results in a quadratic equation, and the largest root is the largest eigenvalue.

Answer 1

The Correct Option is B

Step 1: Set up the characteristic equation. 
To find the eigenvalues of the matrix \( A = \begin{bmatrix} 5 & 4 \\1 & 2 \end{bmatrix} \), solve the characteristic equation: \[ \det(A - \lambda I) = 0, \] where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix. Compute: \[ A - \lambda I = \begin{bmatrix} 5 - \lambda & 4 \\1 & 2 - \lambda \end{bmatrix}. \] The determinant is: \[ \det(A - \lambda I) = (5 - \lambda)(2 - \lambda) - (4)(1) = (5 - \lambda)(2 - \lambda) - 4. \] Expand: \[ (5 - \lambda)(2 - \lambda) = 10 - 5\lambda - 2\lambda + \lambda^2 = \lambda^2 - 7\lambda + 10, \] \[ \det(A - \lambda I) = \lambda^2 - 7\lambda + 10 - 4 = \lambda^2 - 7\lambda + 6. \] Set the determinant to zero: \[ \lambda^2 - 7\lambda + 6 = 0. \] 
Step 2: Solve the characteristic equation. 
Solve the quadratic equation: \[ \lambda^2 - 7\lambda + 6 = 0, \] using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -7 \), \( c = 6 \): \[ \lambda = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 - 24}}{2} = \frac{7 \pm \sqrt{25}}{2} = \frac{7 \pm 5}{2}. \] \[ \lambda_1 = \frac{7 + 5}{2} = \frac{12}{2} = 6, \quad \lambda_2 = \frac{7 - 5}{2} = \frac{2}{2} = 1. \] The eigenvalues are \( \lambda = 6 \) and \( \lambda = 1 \). 
Step 3: Identify the largest eigenvalue. 
The eigenvalues are 6 and 1, so the largest eigenvalue is 6. 
Step 4: Select the correct answer. 
The largest eigenvalue is 6, matching option (2).