Let `A` be a square matrix of order `3` so that sum of elements of each row is `1`. Then the sum elements of matrix `A^(2)` is

Options

A. `1`
B. `3`
C. `0`
D. `6`

Explanation

To solve the problem, we need to find the sum of the elements of the matrix \( A^2 \) given that \( A \) is a \( 3 \times 3 \) matrix where the sum of the elements of each row is 1.

### Step-by-Step Solution:

1. **Define the Matrix \( A \)**:
Let \( A \) be represented as:
\[ A = \begin{pmatrix} a & b & c \\ p & q & r \\ x & y & z \end{pmatrix} \]
Given that the sum of the elements of each row is 1:
\[ a + b + c = 1 \\ p + q + r = 1 \\ x + y + z = 1 \]

2. **Calculate \( A^2 \)**:
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \times A = \begin{pmatrix} a & b & c \\ p & q & r \\ x & y & z \end{pmatrix} \times \begin{pmatrix} a & b & c \\ p & q & r \\ x & y & z \end{pmatrix} \]

The elements of \( A^2 \) can be computed as follows:
- First row:
\[ \begin{pmatrix} a^2 + bp + cx & ab + bq + cy & ac + br + cz \end{pmatrix} \]
- Second row:
\[ \begin{pmatrix} pa + qp + rx & pb + q^2 + ry & pc + qr + rz \end{pmatrix} \]
- Third row:
\[ \begin{pmatrix} xa + yb + zc & xb + yq + zy & xc + yr + z^2 \end{pmatrix} \]

3. **Sum of Elements of \( A^2 \)**:
To find the sum of all elements in \( A^2 \), we sum the elements of each row:
\[ \text{Sum of elements of } A^2 = (a^2 + bp + cx) + (pa + qp + rx) + (xa + yb + zc) + (ab + bq + cy) + (ac + br + cz) + (pb + q^2 + ry) + (pc + qr + rz) + (xb + yq + zy) + (xc + yr + z^2) \]

We can rearrange and group the terms:
\[ = (a + p + x)(a + b + c) + (b + q + y)(a + b + c) + (c + r + z)(a + b + c) \]
Since \( a + b + c = 1 \), we have:
\[ = 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = 3 \]

4. **Conclusion**:
Therefore, the sum of the elements of the matrix \( A^2 \) is:
\[ \boxed{3} \]