If `Aa n dB` are square matrices of the same order and `A` is non-singular, then for a positive integer `n ,(A^(-1)B A)^n` is equal to `A^(-n)B^n A^n` b. `A^n B^n A^(-n)` c. `A^(-1)B^n A^` d. `n(A^(-1)B^A)^`

Options

A. `A^(-n) B^(n) A^(n) `
B. ` A^(n) B^(n) A^(-n) `
C. ` A^(-1) B^(n) A`
D. ` n(A^(-1) B^(A)) `

Answer

To solve the problem, we need to find the expression for \((A^{-1} B A)^n\) where \(A\) and \(B\) are square matrices of the same order, and \(A\) is non-singular. We will proceed step by step.

### Step 1: Understanding the expression
We start with the expression \((A^{-1} B A)^n\). This means we need to multiply \(A^{-1} B A\) by itself \(n\) times.

### Step 2: Calculate for \(n = 1\)
For \(n = 1\):
\[ (A^{-1} B A)^1 = A^{-1} B A \]

### Step 3: Calculate for \(n = 2\)
For \(n = 2\):
\[ (A^{-1} B A)^2 = (A^{-1} B A)(A^{-1} B A) \]
Using the associative property of matrix multiplication:
\[ = A^{-1} B (A A^{-1}) B A \]
Since \(A A^{-1} = I\) (the identity matrix):
\[ = A^{-1} B I B A = A^{-1} B^2 A \]

### Step 4: Calculate for \(n = 3\)
For \(n = 3\):
\[ (A^{-1} B A)^3 = (A^{-1} B A)(A^{-1} B A)(A^{-1} B A) \]
Using the result from \(n = 2\):
\[ = (A^{-1} B^2 A)(A^{-1} B A) \]
Again, applying the associative property:
\[ = A^{-1} B^2 (A A^{-1}) B A = A^{-1} B^2 I B A = A^{-1} B^3 A \]

### Step 5: Generalizing for \(n\)
From the pattern observed:
- For \(n = 1\): \(A^{-1} B^1 A\)
- For \(n = 2\): \(A^{-1} B^2 A\)
- For \(n = 3\): \(A^{-1} B^3 A\)

We can generalize this to:
\[ (A^{-1} B A)^n = A^{-1} B^n A \]

### Conclusion
Thus, the expression \((A^{-1} B A)^n\) simplifies to:
\[ (A^{-1} B A)^n = A^{-1} B^n A \]

### Answer
The correct option is:
**c. \(A^{-1} B^n A\)**