`A=[{:(a,b),(b,-a):}]` and `MA=A^(2m)`, `m in N` for some matrix `M`, then which one of the following is correct ?

Options

A. `M=[{:(a^(2m),b^(2m)),(b^(2m),-a^(2m)):}]`
B. `M=(a^(2)+b^(2))^(m)[{:(1,0),(0,1):}]`
C. `M=(a^(m)+b^(m))[{:(1,0),(0,1):}]`
D. `M=(a^(2)+b^(2))^(m-1)[{:(a,b),(b,-a):}]`

Explanation

To solve the problem, we need to analyze the given matrix \( A \) and the equation \( MA = A^{2m} \) where \( m \in \mathbb{N} \). The matrix \( A \) is defined as:

\[ A = \begin{pmatrix} a & b \\ b & -a \end{pmatrix} \]

### Step 1: Find \( A^2 \)

To find \( A^2 \), we multiply \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} a & b \\ b & -a \end{pmatrix} \cdot \begin{pmatrix} a & b \\ b & -a \end{pmatrix} \]

Calculating the product:

\[ A^2 = \begin{pmatrix} a^2 + b^2 & ab - ab \\ ab - ab & b^2 + a^2 \end{pmatrix} = \begin{pmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{pmatrix} \]

Thus, we can express \( A^2 \) as:

\[ A^2 = (a^2 + b^2) I \]

where \( I \) is the identity matrix.

### Step 2: Find \( A^{2m} \)

Using the result from Step 1, we can find \( A^{2m} \):

\[ A^{2m} = (A^2)^m = ((a^2 + b^2) I)^m = (a^2 + b^2)^m I \]

### Step 3: Use the given equation \( MA = A^{2m} \)

We have:

\[ MA = A^{2m} = (a^2 + b^2)^m I \]

### Step 4: Post-multiply by \( A^{-1} \)

To isolate \( M \), we post-multiply by \( A^{-1} \):

\[ MA A^{-1} = (a^2 + b^2)^m I A^{-1} \]

Since \( AA^{-1} = I \), we simplify to:

\[ M = (a^2 + b^2)^m A^{-1} \]

### Step 5: Find \( A^{-1} \)

To find \( A^{-1} \), we use the formula for the inverse of a 2x2 matrix:

\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]

First, we calculate the determinant of \( A \):

\[ \text{det}(A) = a(-a) - b(b) = -a^2 - b^2 = -(a^2 + b^2) \]

Now, the adjugate of \( A \) is:

\[ \text{adj}(A) = \begin{pmatrix} -a & -b \\ -b & a \end{pmatrix} \]

Thus, we have:

\[ A^{-1} = \frac{1}{-(a^2 + b^2)} \begin{pmatrix} -a & -b \\ -b & a \end{pmatrix} = \frac{1}{a^2 + b^2} \begin{pmatrix} -a & -b \\ -b & a \end{pmatrix} \]

### Step 6: Substitute \( A^{-1} \) back into the equation for \( M \)

Now substituting \( A^{-1} \) into the equation for \( M \):

\[ M = (a^2 + b^2)^m \cdot \frac{1}{a^2 + b^2} \begin{pmatrix} -a & -b \\ -b & a \end{pmatrix} \]

This simplifies to:

\[ M = (a^2 + b^2)^{m-1} \begin{pmatrix} -a & -b \\ -b & a \end{pmatrix} \]

### Conclusion

Thus, the correct expression for \( M \) is:

\[ M = (a^2 + b^2)^{m-1} \begin{pmatrix} -a & -b \\ -b & a \end{pmatrix} \]