The eigenvalues of a matrix are the roots of its characteristic polynomial. The characteristic polynomial of a 2×2 matrix can be computed as follows:
$$|A – \lambda I| = \det \left[ {\begin{array}{{20}{c}} a&b \ c&d \end{array}} – \lambda \begin{array}{{20}{c}} 1&0 \ 0&1 \end{array}} \right] = \lambda ^2 – (a+d) \lambda + ad – bc$$
In this case, $a=2$, $b=-1$, $c=-4$, and $d=5$. Substituting these values into the characteristic polynomial formula, we get:
$$\lambda ^2 – (2+5) \lambda + 2 \cdot 5 – (-1) \cdot (-4) = \lambda ^2 – 7 \lambda + 11$$
Factoring the characteristic polynomial, we get:
$$(\lambda – 2) (\lambda – 5) = 0$$
Therefore, the eigenvalues of the matrix are $\lambda = 2$ and $\lambda = 5$.
To explain each option in brief:
- Option A: $-1$ and $1$. These are not eigenvalues of the matrix, since the characteristic polynomial does not have roots at $-1$ and $1$.
- Option B: $1$ and $6$. These are not eigenvalues of the matrix, since the characteristic polynomial does not have roots at $1$ and $6$.
- Option C: $2$ and $5$. These are the eigenvalues of the matrix, since the characteristic polynomial has roots at $2$ and $5$.
- Option D: $4$ and $-1$. These are not eigenvalues of the matrix, since the characteristic polynomial does not have roots at $4$ and $-1$.