If `A=[[0,1] , [-1,0]]=(alphaI+betaA)^2` then `alpha` and `beta` =

Answer

To solve the equation \( A = (\alpha I + \beta A)^2 \) where \( A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), we will follow these steps:

### Step 1: Write down the matrices
We have:
- \( A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \)
- \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

### Step 2: Compute \( \alpha I \)
Calculating \( \alpha I \):
\[ \alpha I = \alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \alpha & 0 \\ 0 & \alpha \end{bmatrix} \]

### Step 3: Compute \( \beta A \)
Calculating \( \beta A \):
\[ \beta A = \beta \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & \beta \\ -\beta & 0 \end{bmatrix} \]

### Step 4: Add \( \alpha I \) and \( \beta A \)
Now, we add \( \alpha I \) and \( \beta A \):
\[ \alpha I + \beta A = \begin{bmatrix} \alpha & 0 \\ 0 & \alpha \end{bmatrix} + \begin{bmatrix} 0 & \beta \\ -\beta & 0 \end{bmatrix} = \begin{bmatrix} \alpha & \beta \\ -\beta & \alpha \end{bmatrix} \]

### Step 5: Square the result
Next, we need to square the matrix:
\[ (\alpha I + \beta A)^2 = \begin{bmatrix} \alpha & \beta \\ -\beta & \alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ -\beta & \alpha \end{bmatrix} \]

Using matrix multiplication:
\[ = \begin{bmatrix} \alpha^2 + \beta(-\beta) & \alpha\beta + \beta\alpha \\ -\beta\alpha + \alpha(-\beta) & -\beta\beta + \alpha^2 \end{bmatrix} \]
\[ = \begin{bmatrix} \alpha^2 - \beta^2 & 2\alpha\beta \\ -2\alpha\beta & \alpha^2 - \beta^2 \end{bmatrix} \]

### Step 6: Set the squared matrix equal to \( A \)
Now we set the squared matrix equal to \( A \):
\[ \begin{bmatrix} \alpha^2 - \beta^2 & 2\alpha\beta \\ -2\alpha\beta & \alpha^2 - \beta^2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \]

### Step 7: Create equations from the matrix equality
From this equality, we can derive the following equations:
1. \( \alpha^2 - \beta^2 = 0 \)
2. \( 2\alpha\beta = 1 \)

### Step 8: Solve the equations
From the first equation:
\[ \alpha^2 = \beta^2 \implies \alpha = \pm \beta \]

Substituting \( \alpha = \beta \) into the second equation:
\[ 2\alpha^2 = 1 \implies \alpha^2 = \frac{1}{2} \implies \alpha = \pm \frac{1}{\sqrt{2}} \]
Thus, \( \beta = \pm \frac{1}{\sqrt{2}} \).

### Step 9: Conclusion
The values of \( \alpha \) and \( \beta \) are:
\[ \alpha = \pm \frac{1}{\sqrt{2}}, \quad \beta = \pm \frac{1}{\sqrt{2}} \]