Question Bank • QB_EM_ADV_012
NAT • 2 Marks
Section 1: Engineering Mathematics
Linear Algebra
Subtopic: Matrices and determinants
Question Statement:
Consider the $3 \times 3$ upper triangular matrix:
$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix}$
The value of $\det(A^2 - 3A)$ is ________ (answer in integer).
Step-by-Step Solution & Explanation
**Method 1: Eigenvalue Mapping Principle**
The eigenvalues of an upper triangular matrix are simply its diagonal entries:
$\lambda_1 = 1, \quad \lambda_2 = 4, \quad \lambda_3 = 6$
For any polynomial matrix function $f(A) = A^2 - 3A$, the eigenvalues of $f(A)$ are given by $\mu_i = f(\lambda_i) = \lambda_i^2 - 3\lambda_i$:
- $\mu_1 = 1^2 - 3(1) = 1 - 3 = -2$
- $\mu_2 = 4^2 - 3(4) = 16 - 12 = 4$
- $\mu_3 = 6^2 - 3(6) = 36 - 18 = 18$
The determinant of $f(A)$ is the product of its eigenvalues:
$\det(A^2 - 3A) = \mu_1 \cdot \mu_2 \cdot \mu_3 = (-2) \times 4 \times 18 = -144$
**Method 2: Factoring the Matrix Polynomial**
$\det(A^2 - 3A) = \det(A(A - 3I)) = \det(A) \cdot \det(A - 3I)$
$\det(A) = 1 \times 4 \times 6 = 24$
$\det(A - 3I) = (1 - 3)(4 - 3)(6 - 3) = (-2)(1)(3) = -6$
$\det(A^2 - 3A) = 24 \times (-6) = -144$
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