If `A` is a non-singular matrix of order `nxxn` such that `3ABA^(-1)+A=2A^(-1)BA`, then

Options

A. `A` and `B` both are identity matrices
B. `|A+B|=0`
C. `|ABA^(-1)-A^(-1)BA|=0`
D. `A+B` is not a singular matrix

Explanation

To solve the equation \(3ABA^{-1} + A = 2A^{-1}BA\), we will follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ 3ABA^{-1} + A = 2A^{-1}BA \]
We can subtract \(A\) from both sides:
\[ 3ABA^{-1} = 2A^{-1}BA - A \]

### Step 2: Factor Out Common Terms
We can factor out \(A\) from the left side:
\[ 3ABA^{-1} = 2A^{-1}BA - A \]
Now, we can express \(A\) as \(A = I\) (identity matrix) for simplification. However, since \(A\) is non-singular, we will keep it as is for now.

### Step 3: Multiply by \(A^{-1}\)
Next, we multiply both sides by \(A^{-1}\) from the left:
\[ 3B = 2A^{-2}BA - A^{-1} \]

### Step 4: Rearranging Again
Rearranging gives us:
\[ 3B + A^{-1} = 2A^{-2}BA \]

### Step 5: Multiply by \(A\)
Now, we multiply both sides by \(A\):
\[ 3AB + I = 2A^{-1}BA \]

### Step 6: Set \(X = A + B\)
Let \(X = A + B\). Then we can rewrite the equation:
\[ 3A(A + B) = 2A^{-1}(A + B)A \]
This simplifies to:
\[ 3AX = 2A^{-1}XA \]

### Step 7: Taking Determinants
Taking the determinant of both sides:
\[ \text{det}(3AX) = \text{det}(2A^{-1}XA) \]
Using properties of determinants:
\[ 3^n \text{det}(A) \text{det}(X) = 2^n \text{det}(A^{-1}) \text{det}(X) \text{det}(A) \]

### Step 8: Simplifying
Since \(\text{det}(A^{-1}) = \frac{1}{\text{det}(A)}\), we can cancel \(\text{det}(A)\) from both sides:
\[ 3^n \text{det}(X) = 2^n \]

### Step 9: Conclusion
From this, we find:
\[ \text{det}(X) = 0 \Rightarrow \text{det}(A + B) = 0 \]

Thus, the correct answer is that \(\text{det}(A + B) = 0\).