(i)Prove that (adj adjA)-`|A|^(n-2)A`
(ii) Find the value of |adj adj adj A| in terms of |A|

Answer

To solve the given problem, we will break it down into two parts as stated in the question.

### Part (i): Prove that \( \text{adj}(\text{adj} A) = |A|^{n-2} A \)

1. **Start with the property of the adjoint**:
We know that for any square matrix \( A \):
\[ A \cdot \text{adj}(A) = |A| I_n \]
where \( I_n \) is the identity matrix of order \( n \).

2. **Apply the property to \( \text{adj}(A) \)**:
Replace \( A \) with \( \text{adj}(A) \):
\[ \text{adj}(A) \cdot \text{adj}(\text{adj}(A)) = |\text{adj}(A)| I_n \]

3. **Use the property of determinants**:
We know that:
\[ |\text{adj}(A)| = |A|^{n-1} \]
Therefore, we can rewrite the equation:
\[ \text{adj}(A) \cdot \text{adj}(\text{adj}(A)) = |A|^{n-1} I_n \]

4. **Multiply both sides by \( A \)**:
Now, multiply both sides of the equation by \( A \):
\[ A \cdot \text{adj}(A) \cdot \text{adj}(\text{adj}(A)) = A \cdot |A|^{n-1} I_n \]

5. **Substitute \( A \cdot \text{adj}(A) \)**:
From the first property, we know:
\[ A \cdot \text{adj}(A) = |A| I_n \]
Therefore, substituting this into the equation gives:
\[ |A| I_n \cdot \text{adj}(\text{adj}(A)) = |A|^{n} I_n \]

6. **Cancel \( |A| I_n \)**:
Assuming \( |A| \neq 0 \), we can cancel \( |A| I_n \) from both sides:
\[ \text{adj}(\text{adj}(A)) = |A|^{n-2} A \]

Thus, we have proved that:
\[ \text{adj}(\text{adj}(A)) - |A|^{n-2} A = 0 \]
or equivalently,
\[ \text{adj}(\text{adj}(A)) = |A|^{n-2} A \]

### Part (ii): Find the value of \( |\text{adj}(\text{adj}(\text{adj} A))| \) in terms of \( |A| \)

1. **Use the result from part (i)**:
We have already established that:
\[ \text{adj}(\text{adj}(A)) = |A|^{n-2} A \]

2. **Find \( \text{adj}(\text{adj}(\text{adj}(A))) \)**:
Now, we need to find \( \text{adj}(\text{adj}(\text{adj}(A))) \). Using the result from part (i):
\[ \text{adj}(\text{adj}(\text{adj}(A))) = | \text{adj}(A) |^{n-2} \text{adj}(A) \]

3. **Substituting \( | \text{adj}(A) | \)**:
We know that:
\[ | \text{adj}(A) | = |A|^{n-1} \]
Therefore:
\[ \text{adj}(\text{adj}(\text{adj}(A))) = (|A|^{n-1})^{n-2} \text{adj}(A) \]
Simplifying this gives:
\[ \text{adj}(\text{adj}(\text{adj}(A))) = |A|^{(n-1)(n-2)} \text{adj}(A) \]

4. **Finding the determinant**:
Now, we need to find the determinant:
\[ |\text{adj}(\text{adj}(\text{adj}(A)))| = ||A|^{(n-1)(n-2)} \text{adj}(A)| \]
Using the property of determinants:
\[ |\text{adj}(A)| = |A|^{n-1} \]
Therefore:
\[ |\text{adj}(\text{adj}(\text{adj}(A)))| = |A|^{(n-1)(n-2)} |A|^{n-1} = |A|^{(n-1)(n-2) + (n-1)} \]

5. **Final simplification**:
This simplifies to:
\[ |\text{adj}(\text{adj}(\text{adj}(A)))| = |A|^{(n-1)(n-1)} = |A|^{(n-1)^2} \]

Thus, we conclude that:
\[ |\text{adj}(\text{adj}(\text{adj}(A)))| = |A|^{(n-1)^2} \]