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 \( A \)**:
The matrix \( A \) can be represented as:
\[ A = \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & 1 & 1 & \cdots & 1 \\ 1 & 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & 1 \end{pmatrix} \]
where all entries are 1.

2. **Calculate \( A^2 \)**:
To find \( A^2 \), we perform matrix multiplication:
\[ A^2 = A \times A \]
The element in the \( i^{th} \) row and \( j^{th} \) column of \( A^2 \) is calculated as:
\[ (A^2)_{ij} = \sum_{k=1}^{n} A_{ik} \cdot A_{kj} \]
Since all elements of \( A \) are 1, this simplifies to:
\[ (A^2)_{ij} = \sum_{k=1}^{n} 1 \cdot 1 = n \]
for all \( i \) and \( j \).

3. **Resulting Matrix**:
Thus, \( A^2 \) can be represented as:
\[ A^2 = \begin{pmatrix} n & n & n & \cdots & n \\ n & n & n & \cdots & n \\ n & n & n & \cdots & n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n & n & n & \cdots & n \end{pmatrix} \]
which is an \( n \times n \) matrix where every entry is \( n \).

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

### Conclusion:
Thus, we conclude that \( A^2 \) is a scalar multiple of \( A \).