If `A=[[3,-3,4],[2,-3,4],[0,-1,1]]` , then

Options

A. `adj(adjA)=A`
B. `abs(adj(adj(A)))=1`
C. `abs(adj(A))=1`
D. None of these

Explanation

To solve the problem, we need to find the adjoint of the adjoint of matrix \( A \) and verify if it equals \( A \). We will use the formula:

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

where \( n \) is the order of the matrix. For a \( 3 \times 3 \) matrix, \( n = 3 \).

### Step 1: Calculate the Determinant of Matrix \( A \)

Given:
\[ A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \]

We can calculate the determinant \( \det(A) \) using the formula for the determinant of a \( 3 \times 3 \) matrix:

\[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where:
- \( a = 3, b = -3, c = 4 \)
- \( d = 2, e = -3, f = 4 \)
- \( g = 0, h = -1, i = 1 \)

Calculating the determinant step by step:

1. Calculate \( ei - fh \):
\[ (-3)(1) - (4)(-1) = -3 + 4 = 1 \]

2. Calculate \( di - fg \):
\[ (2)(1) - (4)(0) = 2 - 0 = 2 \]

3. Calculate \( dh - eg \):
\[ (2)(-1) - (-3)(0) = -2 - 0 = -2 \]

Now substituting these values into the determinant formula:
\[ \det(A) = 3(1) - (-3)(2) + 4(-2) \]
\[ = 3 + 6 - 8 = 1 \]

### Step 2: Calculate \( \text{adj}(\text{adj}(A)) \)

Using the formula:
\[ \text{adj}(\text{adj}(A)) = (\det A)^{n-2} A \]

Substituting \( \det A = 1 \) and \( n = 3 \):
\[ \text{adj}(\text{adj}(A)) = (1)^{3-2} A = 1 \cdot A = A \]

### Conclusion

Thus, we have shown that:
\[ \text{adj}(\text{adj}(A)) = A \]