Allied & ICAR Exam Bank • ALLIED_1025
NAT • 2 Marks
Section 1: Engineering Mathematics
Linear Algebra
Subtopic: Cayley-Hamilton theorem
Question Statement:
For the matrix $A = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$, using the Cayley-Hamilton theorem, $A^2 - 6A$ equals $k I$. The value of $k$ is:
Step-by-Step Solution & Explanation
Characteristic equation: $\det(A - \lambda I) = (2 - \lambda)(4 - \lambda) - 3 = \lambda^2 - 6\lambda + 8 - 3 = \lambda^2 - 6\lambda + 5 = 0$. By Cayley-Hamilton theorem, $A^2 - 6A + 5I = 0 \implies A^2 - 6A = -5I$. Hence $k = -5$.
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