Question

Consider the matrix $A=\left(\begin{array}{cc}3& -2\\ 4& -1\end{array}\right)$. Then all possible values of $\lambda$ such that the determinant of $B=A-\lambda I$ is $0$, where $I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ and $i=\sqrt{-1}$.

• A. $1\pm 2i$
• B. $2\pm 3i$
• C. $3\pm 4i$
• D. $5\pm 6i$

Answer 1

Answer: (A)

$1\pm 2i$

We have $B=A-\lambda I=\left(\begin{array}{cc}3& -2\\ 4& -1\end{array}\right)-\left(\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right)$
$=\left(\begin{array}{cc}3-\lambda & -2\\ 4& -1-\lambda \end{array}\right)$.
Thus, $det\left(B\right)=(3-\lambda )(-1-\lambda )-(-2)4$
$=-3-3\lambda +\lambda +{\lambda }^2+8$
$={\lambda }^2-2\lambda +5=0$
Solving gives $\lambda =\frac{2\pm 4i}{2}=1\pm 2i$.