`|(1+a_(1),a_2,a_3),(a_1,1+a_2,a_3),(a_1,a_2,1+a_3)|=1+a_1+a_2+a_3`.True or False .

Answer

To determine whether the statement \( |(1 + a_1, a_2, a_3), (a_1, 1 + a_2, a_3), (a_1, a_2, 1 + a_3)| = 1 + a_1 + a_2 + a_3 \) is true or false, we will compute the determinant on the left-hand side (LHS) and check if it equals the right-hand side (RHS).

### Step-by-Step Solution:

1. **Write the Determinant**:
\[ D = \begin{vmatrix} 1 + a_1 & a_2 & a_3 \\ a_1 & 1 + a_2 & a_3 \\ a_1 & a_2 & 1 + a_3 \end{vmatrix} \]

2. **Apply Column Operation**:
We can simplify the determinant by adding all columns together. Let's perform the operation \( C_1 \leftarrow C_1 + C_2 + C_3 \):
\[ D = \begin{vmatrix} (1 + a_1 + a_2 + a_3) & a_2 & a_3 \\ (a_1 + 1 + a_2 + a_3) & 1 + a_2 & a_3 \\ (a_1 + a_2 + 1 + a_3) & a_2 & 1 + a_3 \end{vmatrix} \]

3. **Simplify the Determinant**:
The first column now has the first element as \( 1 + a_1 + a_2 + a_3 \), and the other elements can be simplified:
\[ D = \begin{vmatrix} 1 + a_1 + a_2 + a_3 & a_2 & a_3 \\ 1 + a_2 + a_3 & 1 + a_2 & a_3 \\ 1 + a_1 + a_3 & a_2 & 1 + a_3 \end{vmatrix} \]

4. **Row Operation**:
Now, we can perform a row operation to simplify further. Let's subtract the second row from the first row:
\[ D = \begin{vmatrix} 0 & a_2 & a_3 \\ 1 + a_2 + a_3 & 1 + a_2 & a_3 \\ 1 + a_1 + a_3 & a_2 & 1 + a_3 \end{vmatrix} \]

5. **Expand the Determinant**:
Now we can expand the determinant using the first row:
\[ D = (1 + a_1 + a_2 + a_3) \cdot \begin{vmatrix} 1 + a_2 & a_3 \\ a_2 & 1 + a_3 \end{vmatrix} \]

6. **Calculate the 2x2 Determinant**:
The determinant of the 2x2 matrix is calculated as:
\[ \begin{vmatrix} 1 + a_2 & a_3 \\ a_2 & 1 + a_3 \end{vmatrix} = (1 + a_2)(1 + a_3) - a_2 a_3 = 1 + a_2 + a_3 + a_2 a_3 - a_2 a_3 = 1 + a_2 + a_3 \]

7. **Final Calculation**:
Thus, we have:
\[ D = (1 + a_1 + a_2 + a_3)(1 + a_2 + a_3) \]

8. **Conclusion**:
This shows that \( D = 1 + a_1 + a_2 + a_3 \), confirming that the original statement is true.

### Final Result:
The statement \( |(1 + a_1, a_2, a_3), (a_1, 1 + a_2, a_3), (a_1, a_2, 1 + a_3)| = 1 + a_1 + a_2 + a_3 \) is **True**.