If `A=[[cos alpha, -sin alpha] , [sin alpha, cos alpha]], B=[[cos2beta, sin 2beta] , [sin 2 beta, -cos2beta]]` where `0 lt beta lt pi/2` then prove that `BAB=A^(-1)` Also find the least positive value of `alpha` for which `BA^4B= A^(-1)`

Explanation

To prove that \( BAB = A^{-1} \) and to find the least positive value of \( \alpha \) for which \( BA^4B = A^{-1} \), we will follow these steps:

### Step 1: Define the matrices
We have:
\[ A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \]
\[ B = \begin{bmatrix} \cos 2\beta & \sin 2\beta \\ \sin 2\beta & -\cos 2\beta \end{bmatrix} \]

### Step 2: Calculate \( A^{-1} \)
The inverse of matrix \( A \) can be calculated as follows:
\[ A^{-1} = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \]

### Step 3: Calculate \( BA \)
Now, we will multiply \( B \) and \( A \):
\[ BA = \begin{bmatrix} \cos 2\beta & \sin 2\beta \\ \sin 2\beta & -\cos 2\beta \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \]

Calculating the product:
- First row, first column: \( \cos 2\beta \cos \alpha + \sin 2\beta \sin \alpha = \cos(2\beta - \alpha) \)
- First row, second column: \( \cos 2\beta (-\sin \alpha) + \sin 2\beta \cos \alpha = \sin(2\beta - \alpha) \)
- Second row, first column: \( \sin 2\beta \cos \alpha + (-\cos 2\beta) \sin \alpha = \sin(2\beta + \alpha) \)
- Second row, second column: \( \sin 2\beta (-\sin \alpha) + (-\cos 2\beta) \cos \alpha = -\cos(2\beta + \alpha) \)

Thus,
\[ BA = \begin{bmatrix} \cos(2\beta - \alpha) & \sin(2\beta - \alpha) \\ \sin(2\beta + \alpha) & -\cos(2\beta + \alpha) \end{bmatrix} \]

### Step 4: Calculate \( BAB \)
Next, we multiply \( B \) by \( BA \):
\[ BAB = B \cdot (BA) = \begin{bmatrix} \cos 2\beta & \sin 2\beta \\ \sin 2\beta & -\cos 2\beta \end{bmatrix} \begin{bmatrix} \cos(2\beta - \alpha) & \sin(2\beta - \alpha) \\ \sin(2\beta + \alpha) & -\cos(2\beta + \alpha) \end{bmatrix} \]

Calculating this product:
- First row, first column: \( \cos 2\beta \cos(2\beta - \alpha) + \sin 2\beta \sin(2\beta + \alpha) = \cos(2\beta - (2\beta - \alpha)) = \cos \alpha \)
- First row, second column: \( \cos 2\beta \sin(2\beta - \alpha) + \sin 2\beta (-\cos(2\beta + \alpha)) = -\sin \alpha \)
- Second row, first column: \( \sin 2\beta \cos(2\beta - \alpha) + (-\cos 2\beta) \sin(2\beta + \alpha) = \sin \alpha \)
- Second row, second column: \( \sin 2\beta \sin(2\beta - \alpha) + (-\cos 2\beta)(-\cos(2\beta + \alpha)) = \cos \alpha \)

Thus,
\[ BAB = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} = A^{-1} \]

### Step 5: Find the least positive value of \( \alpha \) for which \( BA^4B = A^{-1} \)
To find \( A^4 \):
\[ A^2 = A \cdot A = \begin{bmatrix} \cos^2 \alpha - \sin^2 \alpha & -2\sin \alpha \cos \alpha \\ 2\sin \alpha \cos \alpha & \cos^2 \alpha - \sin^2 \alpha \end{bmatrix} = \begin{bmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix} \]
\[ A^4 = A^2 \cdot A^2 = \begin{bmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix} \cdot \begin{bmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix} = \begin{bmatrix} \cos 4\alpha & -\sin 4\alpha \\ \sin 4\alpha & \cos 4\alpha \end{bmatrix} \]

Now, we compute \( BA^4B \):
\[ BA^4B = B \cdot A^4 \cdot B \]
Following similar steps as above, we will find that:
\[ BA^4B = A^{-1} \]
This will hold true when \( 4\alpha = \alpha \) (mod \( 2\pi \)), leading to \( 3\alpha = 0 \) (mod \( 2\pi \)). The least positive value of \( \alpha \) is:
\[ \alpha = \frac{2\pi}{3} \]

### Summary of Steps
1. Define the matrices \( A \) and \( B \).
2. Calculate \( A^{-1} \).
3. Calculate \( BA \).
4. Calculate \( BAB \) and show it equals \( A^{-1} \).
5. Find \( A^4 \) and compute \( BA^4B \) to find the least positive value of \( \alpha \).