Allied & ICAR Exam Bank • ALLIED_2619 NAT • 2 Marks

Section 1: Engineering Mathematics

Linear Algebra

Subtopic: Cayley–Hamilton theorem

Question Statement:

Let $A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$. The matrix polynomial $A^3 - 5A^2 + 6A$ evaluates to a zero matrix. The determinant of $A^2 - A$ is:

Answer Key:

12

Step-by-Step Solution & Explanation

The eigenvalues of $A$ are $2$ and $3$. For $f(A) = A^2 - A$, the eigenvalues are $2^2 - 2 = 2$ and $3^2 - 3 = 6$. Thus $\det(A^2 - A) = 2 \times 6 = 12$.
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