Let $A=\begin{pmatrix}\alpha-1 & -1\\6 & \beta\end{pmatrix}$, $\alpha>0$, such that $\det(A)=0$ and $\alpha+\beta=1$. If $I$ denotes the $2\times2$ identity matrix, then the matrix $(I+A)^8$ is:

Step 1: Use given conditions to find $\alpha$ and $\beta$

We are given two conditions: the determinant of matrix $A$ is zero, and the sum of $\alpha$ and $\beta$ is 1. We will use these conditions to find the values of $\alpha$ and $\beta$. First, we calculate the determinant of $A$ and set it to zero. Then, we substitute $\beta = 1-\alpha$ into the determinant equation.

Step 2: Solve for $\alpha$

Substitute $\beta = 1-\alpha$ into the determinant equation. This gives us a quadratic equation in terms of $\alpha$. We simplify and rearrange the terms to get a standard quadratic form.

Step 3: Re-evaluate the determinant calculation

Let's re-evaluate the determinant calculation. The determinant of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad-bc$. So, $\det(A) = (\alpha-1)\beta - (-1)(6) = (\alpha-1)\beta + 6$. Substituting $\beta = 1-\alpha$ into this equation gives us $(\alpha-1)(1-\alpha) + 6 = 0$. Expanding this, we get $\alpha - \alpha^2 - 1 + \alpha + 6 = 0$, which simplifies to $-\alpha^2 + 2\alpha + 5 = 0$, or $\alpha^2 - 2\alpha - 5 = 0$. This is the correct quadratic equation for $\alpha$. However, the problem statement implies a simpler solution for $A$ or $I+A$. Let's check if there's a simpler way to use the given conditions.

Step 4: Re-examine the matrix A and conditions

The problem seems to lead to a quadratic equation for $\alpha$ with irrational roots, which would make $(I+A)^8$ very complex to calculate directly. This suggests there might be a property of the matrix $A$ or $I+A$ that simplifies the calculation. Let's re-examine the given conditions and the structure of matrix $A$.

Step 5: Consider the characteristic equation of A

The characteristic equation of a matrix $A$ is given by $\det(A - \lambda I) = 0$. Expanding this for matrix $A$, we get a quadratic equation in $\lambda$. This equation relates to the eigenvalues of $A$.

Step 6: Simplify the characteristic equation using given conditions

From the characteristic equation, we know that the sum of the diagonal elements (trace) is $\text{Tr}(A) = (\alpha-1) + \beta = (\alpha+\beta) - 1$. The determinant is $\det(A) = (\alpha-1)\beta + 6$. Using the given conditions $\alpha+\beta=1$ and $\det(A)=0$, we substitute these into the characteristic equation. This simplifies to $\lambda^2 - (1-1)\lambda + 0 = 0$, which means $\lambda^2 = 0$.

Step 7: Determine the nature of matrix A and calculate (I+A)^8

Since the characteristic equation is $\lambda^2 = 0$, both eigenvalues of $A$ are 0. This means $A$ is a nilpotent matrix of index 2, i.e., $A^2 = 0$. We can use the binomial expansion for $(I+A)^8$. Since $A^2 = 0$, all higher powers of $A$ ($A^3, A^4, \dots$) will also be zero. Therefore, the expansion simplifies significantly to $I + 8A$.

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