If `A= (1)/(2) ((-1,-sqrt(3)),(sqrt(3),-1))` then `A^(-1)- A^(2)` is equal to a

Options

A. null matrix
B. invertible matrix
C. unit matrix
D. none of these

Answer

To solve the problem, we need to find \( A^{-1} - A^2 \) for the given matrix \( A \).

### Step 1: Define the matrix \( A \)
Given:
\[ A = \frac{1}{2} \begin{pmatrix} -1 & -\sqrt{3} \\ \sqrt{3} & -1 \end{pmatrix} \]
We can simplify this to:
\[ A = \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \]

### Step 2: Calculate the determinant of \( A \)
The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is calculated as:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A \):
- \( a = -\frac{1}{2} \)
- \( b = -\frac{\sqrt{3}}{2} \)
- \( c = \frac{\sqrt{3}}{2} \)
- \( d = -\frac{1}{2} \)

Calculating the determinant:
\[ \text{det}(A) = \left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) - \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) \]
\[ = \frac{1}{4} - \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2} \]

### Step 3: Calculate the inverse of \( A \)
The inverse of a 2x2 matrix is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
Substituting the values:
\[ A^{-1} = \frac{1}{-\frac{1}{2}} \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \]
\[ = -2 \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \]
\[ = \begin{pmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{pmatrix} \]

### Step 4: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \cdot \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \]
Calculating each element:
- First row, first column:
\[ -\frac{1}{2} \cdot -\frac{1}{2} + -\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} - \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2} \]
- First row, second column:
\[ -\frac{1}{2} \cdot -\frac{\sqrt{3}}{2} + -\frac{\sqrt{3}}{2} \cdot -\frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \]
- Second row, first column:
\[ \frac{\sqrt{3}}{2} \cdot -\frac{1}{2} + -\frac{1}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = -\frac{2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2} \]
- Second row, second column:
\[ \frac{\sqrt{3}}{2} \cdot -\frac{\sqrt{3}}{2} + -\frac{1}{2} \cdot -\frac{1}{2} = -\frac{3}{4} + \frac{1}{4} = -\frac{2}{4} = -\frac{1}{2} \]
Thus,
\[ A^2 = \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \]

### Step 5: Calculate \( A^{-1} - A^2 \)
Now we subtract \( A^2 \) from \( A^{-1} \):
\[ A^{-1} - A^2 = \begin{pmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{pmatrix} - \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \]
Calculating each element:
- First row, first column:
\[ 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2} \]
- First row, second column:
\[ -\sqrt{3} - \frac{\sqrt{3}}{2} = -\frac{2\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = -\frac{3\sqrt{3}}{2} \]
- Second row, first column:
\[ \sqrt{3} - \left(-\frac{\sqrt{3}}{2}\right) = \sqrt{3} + \frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]
- Second row, second column:
\[ 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2} \]
Thus,
\[ A^{-1} - A^2 = \begin{pmatrix} \frac{3}{2} & -\frac{3\sqrt{3}}{2} \\ \frac{3\sqrt{3}}{2} & \frac{3}{2} \end{pmatrix} \]

### Final Answer
The result of \( A^{-1} - A^2 \) is:
\[ \begin{pmatrix} \frac{3}{2} & -\frac{3\sqrt{3}}{2} \\ \frac{3\sqrt{3}}{2} & \frac{3}{2} \end{pmatrix} \]