The characteristic equation is \[\det\matrix{-9-\lambda & -24 \\ 4 & 11-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -9-\lambda & -24 \\ 4 & 11-\lambda \end{array} \right\vert &= (-9-\lambda)(11-\lambda)+24\cdot4 \\
&= (-9-\lambda)(11-\lambda)+96 \\ &= \lambda^2-2\,\lambda-3
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-2\,\lambda-3 = (\lambda-3)(\lambda+1) \] So, the eigenvalues are \(\lambda_1 = 3\) and \(\lambda_2 = -1\).

Let \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{-9 & -24 \\ 4 & 11}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=3\vec{v}\), this is, \[(A-3 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-3 I\). We can do this through row reduction of the matrix \[A-3 I = \matrix{-9 & -24 \\ 4 & 11} - \matrix{3 & 0 \\0& 3 }=\matrix{ -12 & -24 \\ 4 & 8}\] This can be done as follows:
\[\begin{aligned}
\matrix{-12&-24\\4&8\\}&\sim\matrix{1&2\\4&8\\}&{\blue{\begin{array}{c}-{{1}\over{12}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&2\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-4R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{-2\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-2\\1}\right\rangle\).

Let \(\lambda = -1\) be an eigenvalue of the matrix \[A=\matrix{-9 & -24 \\ 4 & 11}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-1\vec{v}\), that is, \[(A+I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+I\). We can do this through row reduction of the matrix \[A+I = \matrix{-9 & -24 \\ 4 & 11} - \matrix{-1 & 0 \\0& -1 }=\matrix{ -8 & -24 \\ 4 & 12}\] This can be done as follows:
\[\begin{aligned}
\matrix{-8&-24\\4&12\\}&\sim\matrix{1&3\\4&12\\}&{\blue{\begin{array}{c}-{{1}\over{8}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&3\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-4R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -1\) equals \(\left\{ r \cv{-3\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-3\\1}\right\rangle\).
If possible, we avoided fractions in the solution.