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

We are given:

• \(A=[\begin{matrix}\alpha & −1 \\ 6 & \beta \end{matrix}]\) with \(\alpha >0\).
• \(\det ⁡(A)=0\text{  }⟹\text{  }\alpha \beta −(−1)(6)=0\text{ }\)\(⟹\text{  }\alpha \beta +6=0\text{  }⟹\text{  }\alpha \beta =−6\).
• \(\alpha +\beta =1\text{  }⟹\text{  }\beta =1−\alpha\).

Substitute \(\beta\) in the determinant equation:

\(\alpha (1−\alpha )=−6\) \(\alpha −{\alpha }^{2}=−6\) \({\alpha }^{2}−\alpha −6=0\)

Factoring the quadratic equation:

\((\alpha −3)(\alpha +2)=0\)

Since it's given that \(\alpha >0\), we must have \(\alpha =3\).

Then, \(\beta =1−3=−2\).

Thus, the matrix \(A\) is: \(A=[\begin{matrix}3 & −1 \\ 6 & −2\end{matrix}]\)

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

Since \({A}^{2}=A\), matrix \(A\) is idempotent. .

Using the binomial expansion for matrices that commute (since \(I\) and \(A\) always commute):

\((I+A{)}^{8}={\sum }_{k=0}^{8}(\frac{8}{k}){I}^{8−k}{A}^{k}\)\(=I+{\sum }_{k=1}^{8}(\frac{8}{k}){A}^{k}\)

Because null for all \(k\geq 1\):

\((I+A{)}^{8}=I+({\sum }_{k=1}^{8}(\frac{8}{k}))A\)

We know that \({\sum }_{k=0}^{8}(\frac{8}{k})={2}^{8}=256\),

so \({\sum }_{k=1}^{8}(\frac{8}{k})={2}^{8}−(\frac{8}{0})=256−1=255\).

\((I+A{)}^{8}=I+255A\)

Substitute \(I\) and \(A\) back into the formula:

\((I+A{)}^{8}=[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]+255[\begin{matrix}3 & −1 \\ 6 & −2\end{matrix}]\)

\((I+A{)}^{8}=[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]+[\begin{matrix}765 & −255 \\ 1530 & −510\end{matrix}]\)

\((I+A{)}^{8}=[\begin{matrix}1+765 & −255 \\ 1530 & 1−510\end{matrix}]\)\(=[\begin{matrix}766 & −255 \\ 1530 & −509\end{matrix}]\)