Answer
$$\text {det }A = -17$$
Work Step by Step
Given $$\begin{array}{c}{A=\left[\begin{array}{ccc}
{-2} & {5} \\
{5} & {-4} \end{array}\right] }\end{array} $$
So, we get
$$A=\begin{array}{c}{\left[\begin{array}{ccc}{-2} &{5} \\ {5} & {-4} \end{array}\right] \xrightarrow{R_{2}+\frac{5}{2} R_{1 }} \left[\begin{array}{ccc}{-2} & {5}\\ {0} &{\frac{17}{2}}
\end{array} \right]}
\end{array}$$
\begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=-2 \cdot(\frac{17}{2})=-17}\end{array}
