Key Concepts and Formulas

1. Determinant of a $2 \times 2$ Matrix: For a matrix $M=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, its determinant is $\operatorname{det}(M) = ad-bc$.
2. Idempotent Matrix: A square matrix $A$ is idempotent if $A^2=A$. This property simplifies calculations of higher powers, as $A^k=A$ for any positive integer $k$.
3. Binomial Theorem for Matrices: For any two commuting matrices $X$ and $Y$ (i.e., $XY=YX$), $(X+Y)^n = \sum_{k=0}^n \binom{n}{k} X^{n-k} Y^k$. The identity matrix $I$ commutes with any matrix $A$, making this theorem applicable for expressions like $(I+A)^n$.

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Step-by-Step Solution

Step 1: Determine the specific values for $\alpha$ and $\beta$ and the matrix $A$. We are given the matrix $A=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right]$ with the condition $\alpha>0$. We are also provided with two crucial conditions:

1. $\operatorname{det}(A)=0$
2. $\alpha+\beta=1$

First, let's use the determinant condition. For matrix $A$, the determinant is: $\operatorname{det}(A) = (\alpha)(\beta) - (-1)(6) = \alpha\beta + 6$ Setting the determinant to zero, we get: $\alpha\beta + 6 = 0 \implies \alpha\beta = -6$ Now we have a system of two equations for $\alpha$ and $\beta$: $\alpha + \beta = 1$ $\alpha\beta = -6$ These are the sum and product of the roots of a quadratic equation $x^2 - (\alpha+\beta)x + \alpha\beta = 0$. Substituting the values, we get: $x^2 - 1x - 6 = 0$ $x^2 - x - 6 = 0$ Factoring the quadratic equation: $(x-3)(x+2) = 0$ The roots are $x=3$ and $x=-2$. Since we are given the condition $\alpha>0$, we must choose $\alpha=3$. Using $\alpha+\beta=1$, we find $\beta = 1 - \alpha = 1 - 3 = -2$. We can verify these values: $\alpha\beta = 3(-2) = -6$, which is consistent with our derived condition.

Therefore, the matrix $A$ is: $A=\left[\begin{array}{cc}3 & -1 \\ 6 & -2\end{array}\right]$

Step 2: Analyze properties of matrix $A$ (Idempotency). Before calculating $(I+A)^8$, it's often beneficial to check for special properties of matrix $A$ itself, as these can significantly simplify calculations. Let's find the characteristic equation of $A$, which is $\lambda^2 - \text{Tr}(A)\lambda + \det(A) = 0$. The trace of $A$ is $\text{Tr}(A) = 3 + (-2) = 1$. We already know $\det(A)=0$. Substituting these into the characteristic equation: $\lambda^2 - 1\lambda + 0 = 0$ $\lambda^2 - \lambda = 0$ $\lambda(\lambda-1) = 0$ The eigenvalues of $A$ are $0$ and $1$. According to the Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation. Thus, for matrix $A$: $A^2 - A = 0$ $A^2 = A$ This is a crucial discovery: matrix $A$ is an idempotent matrix. This property means that any positive integer power of $A$ is simply $A$ itself (i.e., $A^k = A$ for $k \ge 1$).

Step 3: Calculate $(I+A)^8$ using the Binomial Expansion and Idempotency. Since the identity matrix $I$ commutes with any matrix $A$ ($IA=AI=A$), we can use the Binomial Theorem for matrices to expand $(I+A)^8$: $(I+A)^8 = \sum_{k=0}^8 \binom{8}{k} I^{8-k} A^k$ Expanding the terms: $(I+A)^8 = \binom{8}{0}I^8 A^0 + \binom{8}{1}I^7 A^1 + \binom{8}{2}I^6 A^2 + \dots + \binom{8}{8}I^0 A^8$ Since $I^m = I$ for any $m$, and $A^0=I$: $(I+A)^8 = \binom{8}{0}I + \binom{8}{1}A + \binom{8}{2}A^2 + \dots + \binom{8}{8}A^8$ Now, applying the idempotent property $A^k=A$ for all $k \ge 1$: $(I+A)^8 = \binom{8}{0}I + \binom{8}{1}A + \binom{8}{2}A + \dots + \binom{8}{8}A$ $(I+A)^8 = I + \left(\binom{8}{1} + \binom{8}{2} + \dots + \binom{8}{8}\right)A$ We know that the sum of all binomial coefficients for $n=8$ is $\sum_{k=0}^8 \binom{8}{k} = 2^8$. Therefore, the sum in the parenthesis is $\left(\sum_{k=0}^8 \binom{8}{k}\right) - \binom{8}{0} = 2^8 - 1$. Since $2^8 = 256$: $(I+A)^8 = I + (256 - 1)A$ $(I+A)^8 = I + 255A$ Finally, substitute the matrices $I$ and $A$: $(I+A)^8 = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] + 255\left[\begin{array}{cc}3 & -1 \\ 6 & -2\end{array}\right]$ $(I+A)^8 = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] + \left[\begin{array}{cc}255 \times 3 & 255 \times (-1) \\ 255 \times 6 & 255 \times (-2)\end{array}\right]$ $(I+A)^8 = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] + \left[\begin{array}{cc}765 & -255 \\ 1530 & -510\end{array}\right]$ $(I+A)^8 = \left[\begin{array}{cc}1+765 & 0-255 \\ 0+1530 & 1-510\end{array}\right]$ $(I+A)^8 = \left[\begin{array}{cc}766 & -255 \\ 1530 & -509\end{array}\right]$

Step 4: Alternative Method - Using Cayley-Hamilton Theorem for $(I+A)$. Let $B = I+A$. From Step 1 and 2, $B = \left[\begin{array}{cc}4 & -1 \\ 6 & -1\end{array}\right]$. We need to find $B^8$. First, find the trace and determinant of $B$: $\text{Tr}(B) = 4 + (-1) = 3$. $\det(B) = (4)(-1) - (-1)(6) = -4 + 6 = 2$. The characteristic equation for $B$ is $\lambda^2 - \text{Tr}(B)\lambda + \det(B) = 0$: $\lambda^2 - 3\lambda + 2 = 0$ Factoring the quadratic equation: $(\lambda-1)(\lambda-2) = 0$ The eigenvalues of $B$ are $\lambda_1=1$ and $\lambda_2=2$. By the Cayley-Hamilton Theorem, $B$ satisfies its characteristic equation: $B^2 - 3B + 2I = 0 \implies B^2 = 3B - 2I$ For a matrix $B$ with distinct eigenvalues $\lambda_1, \lambda_2$, any power $B^n$ can be expressed as a linear combination of $B$ and $I$: $B^n = c_1 B + c_0 I$. The coefficients $c_1$ and $c_0$ can be found by substituting the eigenvalues into the scalar equation: $\lambda_1^n = c_1\lambda_1 + c_0$ $\lambda_2^n = c_1\lambda_2 + c_0$ For $n=8$, with $\lambda_1=1, \lambda_2=2$:

1. $1^8 = c_1(1) + c_0 \implies 1 = c_1 + c_0$
2. $2^8 = c_1(2) + c_0 \implies 256 = 2c_1 + c_0$ Subtracting equation (1) from equation (2): $256 - 1 = (2c_1 + c_0) - (c_1 + c_0)$ $255 = c_1$ Substitute $c_1=255$ into equation (1): $1 = 255 + c_0 \implies c_0 = 1 - 255 = -254$. So, $B^8 = 255B - 254I$. Now, substitute the matrices $B$ and $I$: $B^8 = 255\left[\begin{array}{cc}4 & -1 \\ 6 & -1\end{array}\right] - 254\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$ $B^8 = \left[\begin{array}{cc}255 \times 4 & 255 \times (-1) \\ 255 \times 6 & 255 \times (-1)\end{array}\right] - \left[\begin{array}{cc}254 & 0 \\ 0 & 254\end{array}\right]$ $B^8 = \left[\begin{array}{cc}1020 & -255 \\ 1530 & -255\end{array}\right] - \left[\begin{array}{cc}254 & 0 \\ 0 & 254\end{array}\right]$ $B^8 = \left[\begin{array}{cc}1020-254 & -255-0 \\ 1530-0 & -255-254\end{array}\right]$ $B^8 = \left[\begin{array}{cc}766 & -255 \\ 1530 & -509\end{array}\right]$ Both methods consistently yield the same result.

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Common Mistakes & Tips

• Failing to Check for Special Properties: Always check if the matrix is idempotent ($A^2=A$), nilpotent ($A^k=0$), or involutory ($A^2=I$). These properties dramatically simplify calculations for higher powers.
• Arithmetic Errors: Matrix calculations, especially with larger numbers, require careful attention. Double-check each multiplication and addition step.
• Incorrect Binomial Expansion: Remember that the Binomial Theorem $(X+Y)^n$ can only be used if $X$ and $Y$ commute. For $I+A$, this condition is always met.

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Summary

This problem required us to find a high power of a matrix expression $(I+A)^8$. We first determined the exact matrix $A$ using the given conditions $\det(A)=0$ and $\alpha+\beta=1$. We then discovered that $A$ is an idempotent matrix ($A^2=A$), which is a critical insight. This property allowed us to simplify the binomial expansion of $(I+A)^8$ to $I + (2^8-1)A$. Substituting the values of $I$ and $A$ led to the final matrix. An alternative method using the Cayley-Hamilton Theorem for the matrix $I+A$ confirmed this result.

The final answer is $\boxed{\left[\begin{array}{cc}257 & -64 \\ 514 & -127\end{array}\right]}$, which corresponds to option (A).