`A=[(i,-i),(-i,i)] , A^8[(x),(y)]=[(8),(64)]` has

Options

A. unique solutions
B. no solutions
C. Infinite solutions
D. 2 solutions

Answer

To solve the problem, we need to analyze the given matrix \( A \) and the matrix equation \( A^8 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ 64 \end{pmatrix} \).

### Step 1: Define the Matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} i & -i \\ -i & i \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)
To find \( A^8 \), we first compute \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} i & -i \\ -i & i \end{pmatrix} \cdot \begin{pmatrix} i & -i \\ -i & i \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( i \cdot i + (-i)(-i) = -1 + 1 = 0 \)
- First row, second column: \( i \cdot (-i) + (-i)(i) = -1 - 1 = -2 \)
- Second row, first column: \( -i \cdot i + i(-i) = -1 + 1 = 0 \)
- Second row, second column: \( -i \cdot (-i) + i \cdot i = 1 - 1 = 0 \)

Thus,
\[ A^2 = \begin{pmatrix} 0 & -2 \\ -2 & 0 \end{pmatrix} \]

### Step 3: Calculate \( A^4 \)
Next, we compute \( A^4 \):
\[ A^4 = A^2 \cdot A^2 = \begin{pmatrix} 0 & -2 \\ -2 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & -2 \\ -2 & 0 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 0 \cdot 0 + (-2)(-2) = 4 \)
- First row, second column: \( 0 \cdot (-2) + (-2)(0) = 0 \)
- Second row, first column: \( -2 \cdot 0 + 0 \cdot (-2) = 0 \)
- Second row, second column: \( -2 \cdot (-2) + 0 \cdot 0 = 4 \)

Thus,
\[ A^4 = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} = 4I \]
where \( I \) is the identity matrix.

### Step 4: Calculate \( A^8 \)
Now we can compute \( A^8 \):
\[ A^8 = A^4 \cdot A^4 = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} \cdot \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 16 & 0 \\ 0 & 16 \end{pmatrix} = 16I \]

### Step 5: Set Up the Equation
Now we substitute \( A^8 \) into the equation:
\[ 16I \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ 64 \end{pmatrix} \]
This simplifies to:
\[ \begin{pmatrix} 16x \\ 16y \end{pmatrix} = \begin{pmatrix} 8 \\ 64 \end{pmatrix} \]

### Step 6: Solve for \( x \) and \( y \)
From the equations:
1. \( 16x = 8 \) → \( x = \frac{8}{16} = \frac{1}{2} \)
2. \( 16y = 64 \) → \( y = \frac{64}{16} = 4 \)

### Step 7: Conclusion
The solution is unique, as we have found specific values for \( x \) and \( y \). Therefore, the answer is that there is a unique solution.