If A is a unit matrix of order n, then A(adjA) is

Options

A. row matrix
B. zero matrix
C. unit matrix
D. not unit matrix

Answer

To solve the problem, we need to show that if \( A \) is a unit matrix of order \( n \), then \( A \cdot \text{adj}(A) \) is also a unit matrix of order \( n \).

### Step-by-Step Solution:

1. **Understanding the Unit Matrix**:
A unit matrix (or identity matrix) \( A \) of order \( n \) is defined as a square matrix with 1s on the diagonal and 0s elsewhere. For example, for \( n = 2 \):
\[ A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

2. **Using the Formula for \( A \cdot \text{adj}(A) \)**:
There is a known formula that relates a matrix \( A \) and its adjoint \( \text{adj}(A) \):
\[ A \cdot \text{adj}(A) = \det(A) \cdot I_n \]
where \( \det(A) \) is the determinant of \( A \) and \( I_n \) is the identity matrix of order \( n \).

3. **Calculating the Determinant of \( A \)**:
Since \( A \) is a unit matrix, its determinant is:
\[ \det(A) = 1 \]
This holds true for any unit matrix, regardless of its order.

4. **Substituting the Determinant into the Formula**:
Substituting \( \det(A) = 1 \) into the formula gives:
\[ A \cdot \text{adj}(A) = 1 \cdot I_n = I_n \]

5. **Conclusion**:
Therefore, we conclude that:
\[ A \cdot \text{adj}(A) = I_n \]
This means that \( A \cdot \text{adj}(A) \) is indeed a unit matrix of order \( n \).

### Final Answer:
\( A \cdot \text{adj}(A) \) is a unit matrix of order \( n \).