Let \(\mathrm{A}=\left[\begin{matrix}\alpha & -1 \\ 6 & \beta \end{matrix}\right],\alpha >0\), such that \(\det (A)=0\) …

1. Find \(\alpha\) and \(\beta\) :
Given \(\det (A)=0⟹\alpha \beta -(-1)(6)=0⟹\alpha \beta +6=0⟹\alpha \beta =-6\).
Given \(\alpha +\beta =1⟹\beta =1-\alpha\).
Substitute \(\beta\) into the first equation: \(\alpha (1-\alpha )=-6⟹\alpha -{\alpha }^{2}=-6⟹{\alpha }^{2}-\alpha -6=0\).
Factorizing the quadratic: \((\alpha -3)(\alpha +2)=0\).
Since \(\alpha >0\), we have \(\alpha =3\). Consequently, \(\beta =1-3=-2\).
Therefore, matrix \(A=\left[\begin{matrix}3 & -1 \\ 6 & -2\end{matrix}\right]\).

2. Check properties of A :
Let's compute \({A}^{2}\) :

\({A}^{2}=\left[\begin{matrix}3 & -1 \\ 6 & -2\end{matrix}\right]\left[\begin{matrix}3 & -1 \\ 6 & -2\end{matrix}\right]\\ =\left[\begin{matrix}9-6 & -3+2 \\ 18-12 & -6+4\end{matrix}\right]=\left[\begin{matrix}3 & -1 \\ 6 & -2\end{matrix}\right]=A.\)

Since \({A}^{2}=A\), matrix A is an idempotent matrix. This means \({A}^{n}=A\) for any positive integer n.
3. Compute \((I+A{)}^{8}\) :
Using binomial expansion: \((I+A{)}^{8}=I+\left(\frac{8}{1}\right)A+\left(\frac{8}{2}\right){A}^{2}+⋯+\left(\frac{8}{8}\right){A}^{8}\).
Since \({A}^{k}=A\) for all \(k\geq 1\) :

\((I+A{)}^{8}=I+A\left[\left(\frac{8}{1}\right)+\left(\frac{8}{2}\right)+⋯+\left(\frac{8}{8}\right)\right].\)

The sum of binomial coefficients \(\sum _{k=1}^{n}\left(\frac{n}{k}\right)={2}^{n}-1\).
So, \((I+A{)}^{8}=I+\left({2}^{8}-1\right)A=I+(256-1)A=I+255A\).

\((I+A{)}^{8}=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]+255\left[\begin{matrix}3 & -1 \\ 6 & -2\end{matrix}\right]\\ =\left[\begin{matrix}1+765 & 0-255 \\ 0+1530 & 1-510\end{matrix}\right]=\left[\begin{matrix}766 & -255 \\ 1530 & -509\end{matrix}\right]\)