Key Concepts and Formulas

• Determinant of a Matrix: For a $2 \times 2$ matrix M = \left[ {\matrix{ a & b \cr c & d \cr } } \right], its determinant is $\det(M) = ad - bc$.
• Determinant of Matrix Powers: For any square matrix $M$ and positive integer $n$, $\det(M^n) = (\det(M))^n$.
• Nature of Solutions for a System of Linear Equations ($PX=Q$):
  • If $\det(P) \neq 0$, the system has a unique solution.
  • If $\det(P) = 0$, the system is either inconsistent (no solution) or has infinitely many solutions. To distinguish:
    • No Solution: If Rank($P$) $\neq$ Rank($[P|Q]$), where $[P|Q]$ is the augmented matrix. This means the equations are contradictory.
    • Infinitely Many Solutions: If Rank($P$) = Rank($[P|Q]$) $<$ number of variables. This means the equations are consistent and dependent.
• Fundamental Principle: A system of linear equations can never have exactly two solutions. The possible number of solutions are 0, 1, or infinitely many.

Step-by-Step Solution

Step 1: Calculate the Determinant of Matrix $A$

First, we need to find the determinant of the given matrix $A$. This will help us determine if $A$ is invertible. A = \left[ {\matrix{ i & { - i} \cr { - i} & i \cr } } \right] Using the formula for a $2 \times 2$ determinant: $\det(A) = (i)(i) - (-i)(-i)$ Since $i^2 = -1$: $\det(A) = (-1) - (-1) = -1 + 1 = 0$ Reasoning: A determinant of zero indicates that the matrix $A$ is singular (non-invertible). This is a crucial first step for analyzing the system of equations.

Step 2: Calculate the Determinant of $A^8$

Next, we use the property of determinants that $\det(M^n) = (\det(M))^n$. Applying this to $A^8$: $\det(A^8) = (\det(A))^8$ Substitute the value of $\det(A)$ calculated in Step 1: $\det(A^8) = (0)^8 = 0$ Reasoning: Since $\det(A^8) = 0$, the matrix $A^8$ is also singular. This immediately tells us that the system $A^8 \left[ {\matrix{ x \cr y \cr } } \right] = \left[ {\matrix{ 8 \cr {64} \cr } } \right]$ does not have a unique solution. We must now determine if it has no solution or infinitely many solutions.

Step 3: Calculate the Matrix $A^8$ Explicitly

To analyze the system further, we need the actual form of $A^8$. Let's simplify $A$ and find a pattern for its powers. Observe that $A$ can be written as a scalar multiple of a simpler matrix: A = i \left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right] Let B = \left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]. So, $A = iB$. Now, let's compute powers of $B$: B^2 = B \cdot B = \left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right] \left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right] = \left[ {\matrix{ (1)(1)+(-1)(-1) & (1)(-1)+(-1)(1) \cr (-1)(1)+(1)(-1) & (-1)(-1)+(1)(1) \cr } } \right] B^2 = \left[ {\matrix{ 1+1 & -1-1 \cr -1-1 & 1+1 \cr } } \right] = \left[ {\matrix{ 2 & { - 2} \cr { - 2} & 2 \cr } } \right] We can factor out $2$ from $B^2$: B^2 = 2 \left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right] = 2B This shows a pattern: $B^n = 2^{n-1}B$. Let's verify for $B^3$: $B^3 = B^2 \cdot B = (2B) \cdot B = 2B^2 = 2(2B) = 4B = 2^2 B$. The pattern $B^n = 2^{n-1}B$ is confirmed.

Now we can find $A^8$: $A^8 = (iB)^8 = i^8 B^8$ First, calculate $i^8$: Since $i^4 = (i^2)^2 = (-1)^2 = 1$, then $i^8 = (i^4)^2 = 1^2 = 1$. Next, calculate $B^8$ using the pattern $B^n = 2^{n-1}B$: $B^8 = 2^{8-1}B = 2^7 B = 128B$ Substitute these results back into the expression for $A^8$: $A^8 = 1 \cdot (128B) = 128B$ Finally, substitute the matrix $B$ back: A^8 = 128 \left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right] = \left[ {\matrix{ 128 & { - 128} \cr { - 128} & 128 \cr } } \right] Reasoning: Factoring $i$ from $A$ allowed us to simplify calculations by finding a recursive relation for powers of the real matrix $B$. This is an efficient method for computing high powers of matrices.

Step 4: Analyze the System of Linear Equations

The given system is $A^8 \left[ {\matrix{ x \cr y \cr } } \right] = \left[ {\matrix{ 8 \cr {64} \cr } } \right]$. Substitute the calculated $A^8$: \left[ {\matrix{ 128 & { - 128} \cr { - 128} & 128 \cr } } \right] \left[ {\matrix{ x \cr y \cr } } \right] = \left[ {\matrix{ 8 \cr {64} \cr } } \right] This matrix equation translates into a system of two linear equations:

1. $128x - 128y = 8$
2. $-128x + 128y = 64$

Let's simplify these equations: Divide equation (1) by 128: $x - y = \frac{8}{128} \implies x - y = \frac{1}{16}$ Divide equation (2) by $-128$: $x - y = \frac{64}{-128} \implies x - y = -\frac{1}{2}$ Reasoning: Converting the matrix equation into a standard system of linear equations allows us to easily check for consistency and dependence.

Step 5: Determine the Number of Solutions

We now have the simplified system: $x - y = \frac{1}{16}$ $x - y = -\frac{1}{2}$ These two equations are contradictory because $\frac{1}{16} \neq -\frac{1}{2}$. It is impossible for $x-y$ to simultaneously equal two different values. Therefore, there is no pair of values $(x, y)$ that can satisfy both equations. The system of linear equations is inconsistent.

Thus, based on standard linear algebra principles, the system has no solution.

Common Mistakes & Tips

• Understanding Solution Types: A fundamental error is believing a system of linear equations can have "exactly two solutions". For any linear system, the number of solutions must be 0 (no solution), 1 (unique solution), or infinitely many. There are no other possibilities.
• Determinant for Consistency: Always check the determinant of the coefficient matrix first. If it's non-zero, a unique solution exists. If it's zero, further analysis (like checking the rank of the augmented matrix or simplifying the equations) is required to distinguish between no solution and infinitely many solutions.
• Pattern Recognition: For high powers of matrices, look for patterns ($A^2=kA$, $A^2=kI$, etc.) or try to diagonalize the matrix if possible. Direct multiplication is usually inefficient.

Summary

We began by calculating the determinant of matrix $A$, which was found to be 0. This implied that $A^8$ would also have a determinant of 0, ruling out a unique solution for the system. We then computed $A^8$ by recognizing a pattern in the powers of a simplified matrix related to $A$. This led to A^8 = 128 \left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]. Substituting $A^8$ into the system of equations yielded two contradictory linear equations: $x-y = 1/16$ and $x-y = -1/2$. This inconsistency means that the system has no solution.

The final answer is $\boxed{A}$.