Question Bank • QB_SUBEXP_S1_002 NAT • 2 Marks

Section 1: Engineering Mathematics

Linear Algebra

Subtopic: Eigen values and Eigen vectors

Question Statement:

Consider the matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$. If $\lambda_{max}$ and $\lambda_{min}$ are the maximum and minimum eigenvalues of $A$, the condition number $\kappa(A) = \frac{\lambda_{max}}{\lambda_{min}}$ is ________ (answer in integer).

Answer Key:

3

Step-by-Step Solution & Explanation

The characteristic equation of $A$ is: $\det(A - \lambda I) = (2 - \lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0$ $(\lambda - 3)(\lambda - 1) = 0 \implies \lambda_1 = 3, \lambda_2 = 1$ Maximum eigenvalue $\lambda_{max} = 3$, minimum eigenvalue $\lambda_{min} = 1$. Condition number $\kappa(A) = \frac{\lambda_{max}}{\lambda_{min}} = \frac{3}{1} = 3$.
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