If A is the `n xx n` matrix whose elements are all '1' and B is the `n xx n` matrix whose diagonal elements are all 'n' and other elements are `n-r`, then `A^(2)` is a scalar multiple of

Options

A. B
B. A
C. AB
D. none of these

Answer

To solve the problem, we need to find \( A^2 \) where \( A \) is an \( n \times n \) matrix with all elements equal to 1.

### Step-by-Step Solution:

1. **Define the Matrix \( A \)**:
\[ A = \begin{bmatrix} 1 & 1 & \cdots & 1 \\ 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \end{bmatrix}_{n \times n} \]

2. **Calculate \( A^2 \)**:
To compute \( A^2 \), we perform matrix multiplication \( A \times A \). The element at position \( (i, j) \) in the resulting matrix is calculated by taking the dot product of the \( i \)-th row of the first matrix and the \( j \)-th column of the second matrix.

Each row of \( A \) contains \( n \) ones, and each column of \( A \) also contains \( n \) ones. Therefore, the dot product for any position \( (i, j) \) is:
\[ A^2[i,j] = 1 \times 1 + 1 \times 1 + \cdots + 1 \times 1 = n \]
This is because there are \( n \) terms in the sum, each equal to 1.

3. **Resulting Matrix**:
The resulting matrix \( A^2 \) will be:
\[ A^2 = \begin{bmatrix} n & n & \cdots & n \\ n & n & \cdots & n \\ \vdots & \vdots & \ddots & \vdots \\ n & n & \cdots & n \end{bmatrix}_{n \times n} \]

4. **Expressing \( A^2 \) as a Scalar Multiple**:
We can express \( A^2 \) as:
\[ A^2 = n \cdot A \]
This shows that \( A^2 \) is indeed a scalar multiple of \( A \).

### Conclusion:
Thus, \( A^2 \) is a scalar multiple of the matrix \( A \).