Explanation:

To find the multiplicity of the eigenvalue 0 in the given matrix A, we need to find the number of linearly independent eigenvectors corresponding to the eigenvalue 0.

1. Eigenvectors and eigenvalues:
An eigenvector of a matrix A is a non-zero vector v such that Av = λv, where λ is a scalar called the eigenvalue corresponding to v.

2. Matrix A with entries equal to 1:
Given that A is an n x n matrix with entries equal to 1. Let's assume that n = 3 for the purpose of explaining the concept.

A = | 1 1 1 |
| 1 1 1 |
| 1 1 1 |

3. Eigenvalue equation:
To find the eigenvalues of matrix A, we need to solve the equation (A - λI)v = 0, where I is the identity matrix.

(A - λI)v = 0

For matrix A, we have:

(A - λI) = | 1-λ 1 1 |
| 1 1-λ 1 |
| 1 1 1-λ |

4. Determinant of (A - λI):
To find the eigenvalues, we need to find the determinant of (A - λI) and set it equal to zero.

det(A - λI) = (1-λ)((1-λ)^2 - 1) - (1)((1-λ) - 1) + (1)((1-λ) - 1)
= (1-λ)((1-λ+1)(1-λ-1) - 1) - (1)(1 - λ - 1) + (1)(1 - λ - 1)
= (1-λ)((2-λ)(-λ) - 1) - (1)(-λ) + (1)(-λ)
= (1-λ)(-2λ^2 + 2λ - 1) + λ - λ
= (1-λ)(-2λ^2 + 2λ - 1)
= -2λ^3 + 3λ^2 - 3λ + 1

5. Solving the equation:
To find the eigenvalues, we set the determinant equal to zero and solve for λ:

-2λ^3 + 3λ^2 - 3λ + 1 = 0

By solving this equation, we find that λ = 1 is a repeated eigenvalue.

6. Multiplicity of eigenvalue 0:
Since the eigenvalue 0 does not appear as a solution to the equation -2λ^3 + 3λ^2 - 3λ + 1 = 0, we can conclude that the multiplicity of eigenvalue 0 is 0. Therefore, option 'A' is the correct answer.

Summary:
The matrix A with entries equal to 1 does not have eigenvalue 0. Therefore, the multiplicity of eigenvalue 0 is 0.