Understanding the Problem: Determinant of Adjoint of Adjoint

The question asks us to find the value of the determinant of the adjoint of the adjoint of a given \(3 \times 3\) matrix \(A\). We are given the matrix \(A\) as:

\(A=\left[\begin{array}{rrr} 2 & -1 & 0 \\ -1 & 3 & 0 \\ 1 & 0 & 1 \end{array}\right]\)

To solve this, we need to first calculate the determinant of matrix \(A\), denoted as \( \text{det}(A) \). Then we will use the property relating the determinant of a matrix and the determinant of its adjoint.

Calculating the Determinant of Matrix A

We can calculate the determinant of the \(3 \times 3\) matrix \(A\) by expanding along any row or column. Expanding along the third column is easiest because it contains two zero elements.

\(\text{det}(A) = 2 \cdot \text{Cofactor}(A_{11}) + (-1) \cdot \text{Cofactor}(A_{12}) + 0 \cdot \text{Cofactor}(A_{13}) \)

Or, expanding along the third column:

\(\text{det}(A) = 0 \cdot \text{Cofactor}(A_{13}) + 0 \cdot \text{Cofactor}(A_{23}) + 1 \cdot \text{Cofactor}(A_{33}) \)

We only need to calculate the cofactor of the element \(A_{33}\) which is 1.

The cofactor \( \text{Cofactor}(A_{ij}) = (-1)^{i+j} \cdot \text{Minor}(A_{ij}) \).

The minor \( \text{Minor}(A_{33}) \) is the determinant of the \(2 \times 2\) matrix obtained by removing the 3rd row and 3rd column of \(A\):

\(\text{Minor}(A_{33}) = \text{det}\left[\begin{array}{rr} 2 & -1 \\ -1 & 3 \end{array}\right]\)

\(\text{Minor}(A_{33}) = (2)(3) - (-1)(-1) = 6 - 1 = 5\)

Now, calculate the cofactor \( \text{Cofactor}(A_{33}) \):

\(\text{Cofactor}(A_{33}) = (-1)^{3+3} \cdot \text{Minor}(A_{33}) = (-1)^6 \cdot 5 = 1 \cdot 5 = 5\)

So, the determinant of \(A\) is:

\(\text{det}(A) = 0 \cdot (\text{Cofactor}(A_{13})) + 0 \cdot (\text{Cofactor}(A_{23})) + 1 \cdot 5 = 5\)

Thus, \( \text{det}(A) = 5 \).

Using Properties of Adjoint and Determinant

For an \(n \times n\) matrix \(A\), the determinant of its adjoint is given by the property:

\(\text{det}(\text{adj}(A)) = (\text{det}(A))^{n-1}\)

In this problem, we need to find \( \text{det}(\text{adj}(\text{adj}A)) \). Let's consider \( B = \text{adj}(A) \). Then we need to find \( \text{det}(B) \) where \( B = \text{adj}(A) \). Using the property above for matrix \(B\), we get:

\(\text{det}(\text{adj}(B)) = (\text{det}(B))^{n-1}\)

Substitute \( B = \text{adj}(A) \):

\(\text{det}(\text{adj}(\text{adj}A)) = (\text{det}(\text{adj}(A)))^{n-1}\)

Now, substitute the property for \( \text{det}(\text{adj}(A)) \) into this equation:

\(\text{det}(\text{adj}(\text{adj}A)) = ((\text{det}(A))^{n-1})^{n-1}\)

\(\text{det}(\text{adj}(\text{adj}A)) = (\text{det}(A))^{(n-1)(n-1)} = (\text{det}(A))^{(n-1)^2}\)

For our matrix \(A\), the order is \( n=3 \).

So, \( \text{det}(\text{adj}(\text{adj}A)) = (\text{det}(A))^{(3-1)^2} = (\text{det}(A))^{2^2} = (\text{det}(A))^4 \).

Final Calculation

We found that \( \text{det}(A) = 5 \).

Now, substitute this value into the formula for \( \text{det}(\text{adj}(\text{adj}A)) \):

\(\text{det}(\text{adj}(\text{adj}A)) = (5)^4\)

\((5)^4 = 5 \times 5 \times 5 \times 5 = 25 \times 25 = 625\)

Therefore, the value of \( \text{det}(\text{adj}(\text{adj}A)) \) is 625.

Revision Table: Matrix Determinant and Adjoint Properties

Additional Information: Matrix Adjoint and its Significance

The adjoint of a square matrix \(A\), denoted as \( \text{adj}(A) \), is the transpose of the cofactor matrix of \(A\).

• Cofactor Matrix: A matrix where each element \( (i, j) \) is the cofactor \( \text{Cofactor}(A_{ij}) \).
• Cofactor: \( \text{Cofactor}(A_{ij}) = (-1)^{i+j} \cdot \text{Minor}(A_{ij}) \).
• Minor: \( \text{Minor}(A_{ij}) \) is the determinant of the submatrix formed by deleting the \(i\)-th row and \(j\)-th column of \(A\).
• Significance: The adjoint is closely related to the inverse of a matrix. If \( \text{det}(A) \neq 0 \) (i.e., \(A\) is invertible), then the inverse of \(A\) is given by \( A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \).

Understanding the properties of determinants and adjoints is crucial for solving various problems in linear algebra, including finding matrix inverses, solving systems of linear equations, and analyzing matrix transformations.