In a `4xx4` matrix the sum of each row, column and both the main diagonals is `alpha`. Then the sum of the four corner elements

Options

A. is also `alpha`
B. may not be `alpha`
C. is never equal to `alpha`
D. none of these

Explanation

To solve the problem step by step, we will analyze the properties of the 4x4 matrix and derive the sum of the four corner elements.

### Step-by-Step Solution:

1. **Understanding the Matrix Properties**:
We have a 4x4 matrix where the sum of each row, each column, and both main diagonals is equal to a constant value, denoted as \( \alpha \).

2. **Matrix Representation**:
Let the matrix be represented as follows:
\[ \begin{bmatrix} A_{11} & A_{12} & A_{13} & A_{14} \\ A_{21} & A_{22} & A_{23} & A_{24} \\ A_{31} & A_{32} & A_{33} & A_{34} \\ A_{41} & A_{42} & A_{43} & A_{44} \end{bmatrix} \]

3. **Row and Column Sums**:
From the problem, we know:
- For each row:
\[ A_{i1} + A_{i2} + A_{i3} + A_{i4} = \alpha \quad \text{for } i = 1, 2, 3, 4 \]
- For each column:
\[ A_{1j} + A_{2j} + A_{3j} + A_{4j} = \alpha \quad \text{for } j = 1, 2, 3, 4 \]

4. **Diagonal Sums**:
The sums of the main diagonals are also equal to \( \alpha \):
- First diagonal:
\[ A_{11} + A_{22} + A_{33} + A_{44} = \alpha \]
- Second diagonal:
\[ A_{14} + A_{23} + A_{32} + A_{41} = \alpha \]

5. **Finding the Sum of Corner Elements**:
We need to find the sum of the four corner elements:
\[ S = A_{11} + A_{14} + A_{41} + A_{44} \]

6. **Using the Row and Column Sums**:
We can express the total sum of the matrix using the row sums:
- The total sum of all rows (4 rows) is \( 4\alpha \).
- The total sum of all columns (4 columns) is also \( 4\alpha \).

7. **Combining the Sums**:
We can add the equations for the first row and the first column:
\[ (A_{11} + A_{12} + A_{13} + A_{14}) + (A_{11} + A_{21} + A_{31} + A_{41}) = \alpha + \alpha \]
This simplifies to:
\[ 2A_{11} + A_{12} + A_{13} + A_{14} + A_{21} + A_{31} + A_{41} = 2\alpha \]

8. **Using the Diagonal Sums**:
Adding both diagonal equations:
\[ (A_{11} + A_{22} + A_{33} + A_{44}) + (A_{14} + A_{23} + A_{32} + A_{41}) = \alpha + \alpha \]
This simplifies to:
\[ A_{11} + A_{14} + A_{41} + A_{44} + A_{22} + A_{33} + A_{23} + A_{32} = 2\alpha \]

9. **Eliminating Middle Elements**:
We can rearrange the equations to isolate \( S \):
\[ S + (A_{22} + A_{33} + A_{12} + A_{13} + A_{21} + A_{31}) = 2\alpha \]
Since the sum of the middle elements (rows and columns) also equals \( 2\alpha \), we can conclude:
\[ S = \alpha \]

10. **Final Result**:
Therefore, the sum of the four corner elements is:
\[ S = A_{11} + A_{14} + A_{41} + A_{44} = \alpha \]

### Conclusion:
The answer is \( \alpha \), which corresponds to Option A.