If A be a square matrix of order 3, such that `|A|=sqrt5`, then `|Adj(-3A^(-2))|` is equal to

Answer

To solve the problem, we need to find the determinant of the adjugate of the matrix \(-3A^{-2}\), given that the determinant of matrix \(A\) is \(|A| = \sqrt{5}\).

### Step-by-Step Solution:

1. **Understand the properties of determinants**:
The determinant of the adjugate of a matrix \(B\) is given by the formula:
\[ |Adj(B)| = |B|^{n-1} \]
where \(n\) is the order of the matrix. Since \(A\) is a \(3 \times 3\) matrix, \(n = 3\).

2. **Calculate the determinant of \(-3A^{-2}\)**:
Using the property of determinants, we have:
\[ |kB| = k^n |B| \]
where \(k\) is a scalar and \(B\) is a matrix. Here, \(k = -3\) and \(B = A^{-2}\). Thus,
\[ |-3A^{-2}| = (-3)^3 |A^{-2}| \]
\[ = -27 |A^{-2}| \]

3. **Calculate \(|A^{-2}|\)**:
The determinant of the inverse of a matrix is given by:
\[ |A^{-1}| = \frac{1}{|A|} \]
Therefore,
\[ |A^{-2}| = |A^{-1}|^2 = \left(\frac{1}{|A|}\right)^2 = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5} \]

4. **Substitute back to find \(|-3A^{-2}|\)**:
Now substituting \(|A^{-2}|\) back into the determinant of \(-3A^{-2}\):
\[ |-3A^{-2}| = -27 \cdot \frac{1}{5} = -\frac{27}{5} \]

5. **Find the determinant of the adjugate**:
Now, we can use the formula for the determinant of the adjugate:
\[ |Adj(-3A^{-2})| = |-3A^{-2}|^{3-1} = |-3A^{-2}|^2 \]
Thus,
\[ |Adj(-3A^{-2})| = \left(-\frac{27}{5}\right)^2 = \frac{729}{25} \]

### Final Answer:
\[ |Adj(-3A^{-2})| = \frac{729}{25} \]