`A` and `B` be `3xx3` matrices such that `AB+A+B=0`, then

Options

A. `(A+B)^(2)=A^(2)+2AB+B^(2)`
B. `|A|=|B|`
C. `A^(2)=B^(2)`
D. none of these

Explanation

To solve the problem given that \( A \) and \( B \) are \( 3 \times 3 \) matrices such that \( AB + A + B = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ AB + A + B = 0 \]
Rearranging gives us:
\[ AB + A + B + I = I \]

### Step 2: Factoring the Left Side
Now, we can factor the left-hand side:
\[ AB + A + B + I = (A + I)(B + I) \]
Thus, we have:
\[ (A + I)(B + I) = I \]

### Step 3: Inverses
From the equation \( (A + I)(B + I) = I \), we can conclude that \( A + I \) and \( B + I \) are inverses of each other. This implies:
\[ A + I = (B + I)^{-1} \]
and
\[ B + I = (A + I)^{-1} \]

### Step 4: Multiplying Both Sides
Now, we can multiply both sides of the equation \( (A + I)(B + I) = I \) in a different order:
\[ (A + I)(B + I) = (B + I)(A + I) \]
Expanding both sides:
\[ AB + A + B + I = BA + B + A + I \]

### Step 5: Simplifying
We can cancel \( A \), \( B \), and \( I \) from both sides:
\[ AB = BA \]
This shows that \( A \) and \( B \) commute.

### Step 6: Checking the First Option
Now, we check the first option:
\[ (A + B)^2 = A^2 + 2AB + B^2 \]
Expanding \( (A + B)^2 \):
\[ (A + B)(A + B) = A^2 + AB + BA + B^2 \]
Since \( AB = BA \), we can write:
\[ A^2 + AB + AB + B^2 = A^2 + 2AB + B^2 \]
Thus, the first option is verified:
\[ (A + B)^2 = A^2 + 2AB + B^2 \]

### Conclusion
The first option is correct. Therefore, the answer to the question is:
\[ \text{Option 1: } A + B \text{ whole square is equal to } A^2 + 2AB + B^2 \]