Discussion
For the matrix A = \(\begin{bmatrix} 5 & 3 \\ 1 & 3 \end{bmatrix}\), ONE of the normalized eigen vectors is given as
Answer: B
A normalized eigenvector is \(\left(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}\right)^{T}\).
The characteristic equation gives the eigenvalues:
\[ \begin{aligned} (5-\lambda)(3-\lambda)-3 &= 0 \\ \lambda^{2}-8\lambda+12 &= 0 \\ \lambda &= 2,\ 6 \end{aligned} \]
For \(\lambda = 2\): \(3x+3y=0\), so \(y=-x\) and the eigenvector is (1, -1), normalized to \((1/\sqrt{2}, -1/\sqrt{2})\). For \(\lambda = 6\): \(-x+3y=0\), giving (3, 1), normalized to \((3/\sqrt{10}, 1/\sqrt{10})\).
Option C has the right magnitudes for the second eigenvector but the wrong sign on its second component; check: \(A(3,-1)^{T} = (12, 0)^{T}\), not a multiple of (3, -1). Options A and D are unit vectors but not eigenvectors of A.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development