Topic: Engineering Mathematics
The smallest Eigen value of the matrix $A=\begin{pmatrix}4 & 3 & 1\\0 & 6 & 0\\0 & 4 & 7\end{pmatrix}$ is
- A. 1
- B. 3
- C. 4
- D. 6
Answer: C. 4
Explanation: The characteristic equation is given by $\det(A - \lambda I) = 0$.
$\begin{vmatrix} 4-\lambda & 3 & 1 \\\ 0 & 6-\lambda & 0 \\\ 0 & 4 & 7-\lambda \end{vmatrix} = 0$.
Expanding along the first column:
$(4-\lambda) \begin{vmatrix} 6-\lambda & 0 \\\ 4 & 7-\lambda \end{vmatrix} = (4-\lambda)(6-\lambda)(7-\lambda) = 0$.
The eigenvalues are $\lambda = 4, 6, 7$.
The smallest eigenvalue is 4.