If `A=[{:(3,-3,4),(2,-3,4),(0,-1,1):}]` , then the trace of the matrix `Adj(AdjA)` is

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To find the trace of the matrix \( \text{Adj}(\text{Adj}(A)) \), we will follow these steps:

### Step 1: Define the matrix \( A \)

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

### Step 2: Calculate the determinant of \( A \)

To find the determinant of \( A \), we can use the formula for the determinant of a \( 3 \times 3 \) matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is represented as:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

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

Calculating the determinant:
\[ \text{det}(A) = 3((-3)(1) - (4)(-1)) - (-3)((2)(1) - (4)(0)) + 4((2)(-1) - (-3)(0)) \]
\[ = 3(-3 + 4) + 3(2) + 4(-2) \]
\[ = 3(1) + 6 - 8 \]
\[ = 3 + 6 - 8 = 1 \]

### Step 3: Find \( \text{Adj}(\text{Adj}(A)) \)

The property of the adjoint states that:
\[ \text{Adj}(\text{Adj}(A)) = \text{det}(A)^{n-2} A \]
where \( n \) is the order of the matrix. For our \( 3 \times 3 \) matrix, \( n = 3 \).

Thus,
\[ \text{Adj}(\text{Adj}(A)) = \text{det}(A)^{3-2} A = \text{det}(A) A = 1 \cdot A = A \]

### Step 4: Calculate the trace of \( \text{Adj}(\text{Adj}(A)) \)

The trace of a matrix is the sum of its diagonal elements. Therefore, we compute the trace of \( A \):
\[ \text{Trace}(A) = 3 + (-3) + 1 = 1 \]

### Conclusion

The trace of the matrix \( \text{Adj}(\text{Adj}(A)) \) is:
\[ \text{Trace}(\text{Adj}(\text{Adj}(A))) = 1 \]