If U=`[((1)/(sqrt2),(-1)/(sqrt2)),((1)/(sqrt2),(1)/(sqrt2))]`, then `U^(-1)` is

Options

A. `U^(T)`
B. U
C. I
D. 0

Answer

To find the inverse of the matrix \( U \), we can follow these steps:

Given:
\[ U = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \]

### Step 1: Calculate the Determinant of \( U \)

The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \).

For our matrix \( U \):
- \( a = \frac{1}{\sqrt{2}} \)
- \( b = -\frac{1}{\sqrt{2}} \)
- \( c = \frac{1}{\sqrt{2}} \)
- \( d = \frac{1}{\sqrt{2}} \)

Calculating the determinant:
\[ \text{det}(U) = \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\right) \]
\[ = \frac{1}{2} + \frac{1}{2} = 1 \]

### Step 2: Check if the Inverse Exists

Since the determinant \( \text{det}(U) = 1 \) (which is not equal to 0), the inverse of \( U \) exists.

### Step 3: Find the Adjoint of \( U \)

For a \( 2 \times 2 \) matrix, the adjoint is found by swapping the elements on the main diagonal and changing the signs of the other two elements.

The adjoint of \( U \) is:
\[ \text{adj}(U) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \]

### Step 4: Calculate the Inverse of \( U \)

The inverse of \( U \) is given by the formula:
\[ U^{-1} = \frac{1}{\text{det}(U)} \cdot \text{adj}(U) \]
Substituting the values:
\[ U^{-1} = \frac{1}{1} \cdot \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \]
\[ = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \]

### Final Result

Thus, the inverse of matrix \( U \) is:
\[ U^{-1} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \]