If A is a 3 x 3 matrix such that |5-adjA|=5 , then |A| is equal to :

Options

A. `pm1/5`
B. `pm5`
C. `pm1`
D. `pm1/25`

Answer

To solve the problem, we need to find the determinant of the matrix \( A \) given that \( |5 \cdot \text{adj} A| = 5 \).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We are given that \( |5 \cdot \text{adj} A| = 5 \).

2. **Using the Property of Determinants:**
The property of determinants states that for any scalar \( k \) and an \( n \times n \) matrix \( A \),
\[ |kA| = k^n |A| \]
For our case, since \( A \) is a \( 3 \times 3 \) matrix (\( n = 3 \)), we can write:
\[ |5 \cdot \text{adj} A| = 5^3 |\text{adj} A| = 125 |\text{adj} A| \]

3. **Setting Up the Equation:**
From the given information, we have:
\[ 125 |\text{adj} A| = 5 \]
Dividing both sides by 125 gives:
\[ |\text{adj} A| = \frac{5}{125} = \frac{1}{25} \]

4. **Using the Determinant of the Adjoint:**
The determinant of the adjoint of a matrix \( A \) is related to the determinant of \( A \) by the formula:
\[ |\text{adj} A| = |A|^{n-1} \]
For a \( 3 \times 3 \) matrix, this becomes:
\[ |\text{adj} A| = |A|^{3-1} = |A|^2 \]

5. **Setting Up the Equation for Determinant of A:**
We now have:
\[ |A|^2 = \frac{1}{25} \]

6. **Solving for |A|:**
Taking the square root of both sides gives:
\[ |A| = \pm \sqrt{\frac{1}{25}} = \pm \frac{1}{5} \]

7. **Conclusion:**
Therefore, the determinant of matrix \( A \) is:
\[ |A| = \pm \frac{1}{5} \]

### Final Answer:
Thus, \( |A| \) is equal to \( \pm \frac{1}{5} \).