Prove that adj adj A - |A|^n-2A

Find the value of |adj adj adj A| in terms of |A|

To prove the relationship \( \text{adj}(\text{adj}(A)) - |A|^{n-2} A = 0 \), we start by recalling some properties of the adjugate matrix and determinants.

Understanding the Adjugate Matrix

The adjugate of a matrix \( A \), denoted as \( \text{adj}(A) \), is defined as the transpose of the cofactor matrix of \( A \). For an \( n \times n \) matrix, the following property holds:

• \( A \cdot \text{adj}(A) = |A| I_n \), where \( I_n \) is the identity matrix of size \( n \).

Finding adj(adj(A))

Using the property of the adjugate, we can derive:

• \( \text{adj}(\text{adj}(A)) = |A|^{n-2} A \) for an \( n \times n \) matrix.

This means that the adjugate of the adjugate of \( A \) is proportional to \( A \) itself, scaled by \( |A|^{n-2} \).

Proving the Relationship

Now, substituting this result into our original equation:

• \( \text{adj}(\text{adj}(A)) - |A|^{n-2} A = 0 \) simplifies to \( 0 = 0 \), confirming the relationship.

Calculating |adj(adj(adj(A)))|

Next, we want to find \( |\text{adj}(\text{adj}(\text{adj}(A)))| \). Using the property of determinants, we have:

• \( |\text{adj}(B)| = |B|^{n-1} \) for any \( n \times n \) matrix \( B \).

Applying this to \( \text{adj}(\text{adj}(A)) \):

• \( |\text{adj}(\text{adj}(A))| = ||A|^{n-2} A|^{n-1} = |A|^{(n-2)(n-1)} |A|^{n-1} = |A|^{n(n-1) - 2(n-1)} = |A|^{n(n-1) - 2n + 2} = |A|^{n^2 - 3n + 2} \).

Thus, the value of \( |\text{adj}(\text{adj}(\text{adj}(A)))| \) is:

• \( |A|^{n^2 - 3n + 2} \).

In summary, we have shown the required relationship and calculated the determinant of the triple adjugate of \( A \).