A) \[1\pm 2\,i\]
B) \[2\pm 3\,i\]
C) \[3\pm 4\,i\]
D) \[5\pm 6\,i\]
Correct Answer: C
Solution :
We have \[B=A-\lambda I\,=\left( \begin{matrix} 3 & -2 \\ 4 & -1 \\ \end{matrix} \right)\,-\left( \begin{matrix} \lambda & 0 \\ 0 & \lambda \\ \end{matrix} \right)\] \[=\left( \begin{matrix} 3-\lambda & -2 \\ 4 & -1-\lambda \\ \end{matrix} \right)\] Thus det \[=(3-\lambda )\,(-1-\lambda )\,-(-2)\,4=-3\] \[-3\lambda +\lambda +{{\lambda }^{2}}+8\] \[={{\lambda }^{2}}\,-2\lambda +5=0\] Solving gives \[\lambda =\frac{2\pm \,4i}{2}\,=1\pm \,2i\]
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