1. Given matrix and conditions

We have

$$A=\begin{bmatrix}\alpha & -1\\ 6 & \beta\end{bmatrix}, \qquad \alpha>0$$

with

$$\det(A)=0, \qquad \alpha+\beta=1.$$

Now,

$$\det(A)=\alpha\beta-(-1)(6)=\alpha\beta+6=0$$

so

$$\alpha\beta=-6.$$

Also,

$$\alpha+\beta=1.$$

Thus $\alpha,\beta$ are roots of

$$t^2-(\alpha+\beta)t+\alpha\beta=0$$

that is,

$$t^2-t-6=0.$$

Factorizing,

$$(t-3)(t+2)=0.$$

So the roots are $3$ and $-2$. Since $\alpha>0$, we get

$$\alpha=3, \qquad \beta=-2.$$

Hence,

$$A=\begin{bmatrix}3 & -1\\ 6 & -2\end{bmatrix}.$$

2. Find $I+A$

The identity matrix is

$$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$

Therefore,

$$I+A=\begin{bmatrix}4 & -1\\ 6 & -1\end{bmatrix}.$$

Let

$$M=I+A=\begin{bmatrix}4 & -1\\ 6 & -1\end{bmatrix}.$$

3. Look for a simple power pattern

Compute $M^2$:

$$M^2= \begin{bmatrix}4 & -1\\ 6 & -1\end{bmatrix} \begin{bmatrix}4 & -1\\ 6 & -1\end{bmatrix}.$$

Now multiply:

$$M^2= \begin{bmatrix} 4\cdot 4+(-1)\cdot 6 & 4\cdot(-1)+(-1)\cdot(-1)\\ 6\cdot 4+(-1)\cdot 6 & 6\cdot(-1)+(-1)\cdot(-1) \end{bmatrix} = \begin{bmatrix} 16-6 & -4+1\\ 24-6 & -6+1 \end{bmatrix} = \begin{bmatrix}10 & -3\\18 & -5\end{bmatrix}.$$

Compare this with $M$:

$$2M=2\begin{bmatrix}4 & -1\\ 6 & -1\end{bmatrix}= \begin{bmatrix}8 & -2\\ 12 & -2\end{bmatrix},$$

so that is not directly useful.

Let us instead use the characteristic polynomial of $M$.

4. Characteristic polynomial of $M$

For

$$M=\begin{bmatrix}4 & -1\\ 6 & -1\end{bmatrix},$$

its trace is

$$\operatorname{tr}(M)=4+(-1)=3$$

and determinant is

$$\det(M)=4(-1)-(-1)(6)=-4+6=2.$$

Therefore the characteristic polynomial is

$$\lambda^2-3\lambda+2=0$$

which factors as

$$(\lambda-1)(\lambda-2)=0.$$

So eigenvalues are $1$ and $2$.

By Cayley-Hamilton,

$$M^2-3M+2I=0.$$

This implies powers of $M$ can be expressed linearly in $M$ and $I$.

Since eigenvalues are $1,2$, for any polynomial expression we can interpolate:

$$M^8=aM+bI.$$

For eigenvalue $1$:

$$1^8=a(1)+b \implies a+b=1.$$

For eigenvalue $2$:

$$2^8=256=2a+b.$$

Subtracting,

$$a=255,$$

then

$$b=1-255=-254.$$

So,

$$M^8=255M-254I.$$

5. Compute $M^8$

Now,

$$255M=255\begin{bmatrix}4 & -1\\ 6 & -1\end{bmatrix} = \begin{bmatrix}1020 & -255\\ 1530 & -255\end{bmatrix}.$$

Also,

$$254I=\begin{bmatrix}254 & 0\\ 0 & 254\end{bmatrix}.$$

Hence,

$$M^8=255M-254I = \begin{bmatrix}1020-254 & -255\\ 1530 & -255-254\end{bmatrix} = \begin{bmatrix}766 & -255\\ 1530 & -509\end{bmatrix}.$$

Therefore,

$$(I+A)^8=\begin{bmatrix}766 & -255\\ 1530 & -509\end{bmatrix}.$$

6. Compare with options

This matches Option B.

7. Verification with stored answer

Stored correct answer is B, which agrees with our result.