A and B are two non-singular square matrices of each `3xx3` such that AB = A and BA = B and `|A+B| ne 0` then

Options

A. `|A+B|=0`
B. `|A+B|=8`
C. `|A-B|=1`
D. `|A+B|=2`

Answer

To solve the problem, we need to analyze the given conditions about the matrices A and B. Let's break it down step by step.

### Step 1: Understand the given conditions
We are given two non-singular \(3 \times 3\) matrices \(A\) and \(B\) such that:
1. \(AB = A\)
2. \(BA = B\)
3. \(|A + B| \neq 0\)

Since both \(A\) and \(B\) are non-singular, we know that \(|A| \neq 0\) and \(|B| \neq 0\).

**Hint:** Non-singular matrices have non-zero determinants.

### Step 2: Analyze the equation \(AB = A\)
From the equation \(AB = A\), we can manipulate it as follows:
\[ AB - A = 0 \implies A(B - I) = 0 \]
Since \(A\) is non-singular, we can conclude that:
\[ B - I = 0 \implies B = I \]

**Hint:** If \(A\) is non-singular, then the only solution to \(A \cdot X = 0\) is \(X = 0\).

### Step 3: Analyze the equation \(BA = B\)
Now, let's analyze the second equation \(BA = B\):
\[ BA - B = 0 \implies B(A - I) = 0 \]
Again, since \(B\) is non-singular, we can conclude that:
\[ A - I = 0 \implies A = I \]

**Hint:** Similar to the previous step, if \(B\) is non-singular, then the only solution to \(B \cdot Y = 0\) is \(Y = 0\).

### Step 4: Determine \(A + B\)
Now that we have found:
\[ A = I \quad \text{and} \quad B = I \]
We can substitute these values into \(A + B\):
\[ A + B = I + I = 2I \]

**Hint:** The identity matrix \(I\) has a determinant of 1.

### Step 5: Calculate the determinant \(|A + B|\)
Now we need to find the determinant:
\[ |A + B| = |2I| = 2^3 |I| = 2^3 \cdot 1 = 8 \]

**Hint:** The determinant of a scalar multiple of a matrix is the scalar raised to the power of the matrix's order multiplied by the determinant of the matrix.

### Conclusion
Thus, we find that:
\[ |A + B| = 8 \]

This satisfies the condition \(|A + B| \neq 0\).

### Final Answer
The final result is:
\[ |A + B| = 8 \]