If `A=[(2, 1,-1),(3, 5,2),(1, 6, 1)]`, then `tr(Aadj(adjA))` is equal to (where, tr (P) denotes the trace of the matrix P i.e. the sum of all the diagonal elements of the matrix P and adj(P) denotes the adjoint of matrix P)

Options

A. 7
B. 18
C. `-58`
D. `-1624`

Answer

To solve the problem, we need to find the value of \( \text{tr}(A \cdot \text{adj}(\text{adj}(A))) \) where \( A = \begin{pmatrix} 2 & 1 & -1 \\ 3 & 5 & 2 \\ 1 & 6 & 1 \end{pmatrix} \).

### Step 1: Use the property of adjoint
The property of the adjoint states that:
\[ \text{adj}(\text{adj}(A)) = \det(A) \cdot A \]
Thus, we can rewrite our expression:
\[ \text{tr}(A \cdot \text{adj}(\text{adj}(A))) = \text{tr}(A \cdot (\det(A) \cdot A)) = \det(A) \cdot \text{tr}(A^2) \]

### Step 2: Calculate \( \det(A) \)
To find \( \det(A) \), we can use the formula for the determinant of a 3x3 matrix:
\[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ A = \begin{pmatrix} 2 & 1 & -1 \\ 3 & 5 & 2 \\ 1 & 6 & 1 \end{pmatrix} \]
Calculating the determinant:
\[ \det(A) = 2(5 \cdot 1 - 2 \cdot 6) - 1(3 \cdot 1 - 2 \cdot 1) - 1(3 \cdot 6 - 5 \cdot 1) \]
Calculating each term:
\[ = 2(5 - 12) - 1(3 - 2) - 1(18 - 5) \]
\[ = 2(-7) - 1(1) - 1(13) \]
\[ = -14 - 1 - 13 = -28 \]
Thus, \( \det(A) = -28 \).

### Step 3: Calculate \( A^2 \)
Next, we need to calculate \( A^2 = A \cdot A \):
\[ A^2 = \begin{pmatrix} 2 & 1 & -1 \\ 3 & 5 & 2 \\ 1 & 6 & 1 \end{pmatrix} \cdot \begin{pmatrix} 2 & 1 & -1 \\ 3 & 5 & 2 \\ 1 & 6 & 1 \end{pmatrix} \]
Calculating the elements:
- First row:
- \( 2 \cdot 2 + 1 \cdot 3 + (-1) \cdot 1 = 4 + 3 - 1 = 6 \)
- \( 2 \cdot 1 + 1 \cdot 5 + (-1) \cdot 6 = 2 + 5 - 6 = 1 \)
- \( 2 \cdot (-1) + 1 \cdot 2 + (-1) \cdot 1 = -2 + 2 - 1 = -1 \)

- Second row:
- \( 3 \cdot 2 + 5 \cdot 3 + 2 \cdot 1 = 6 + 15 + 2 = 23 \)
- \( 3 \cdot 1 + 5 \cdot 5 + 2 \cdot 6 = 3 + 25 + 12 = 40 \)
- \( 3 \cdot (-1) + 5 \cdot 2 + 2 \cdot 1 = -3 + 10 + 2 = 9 \)

- Third row:
- \( 1 \cdot 2 + 6 \cdot 3 + 1 \cdot 1 = 2 + 18 + 1 = 21 \)
- \( 1 \cdot 1 + 6 \cdot 5 + 1 \cdot 6 = 1 + 30 + 6 = 37 \)
- \( 1 \cdot (-1) + 6 \cdot 2 + 1 \cdot 1 = -1 + 12 + 1 = 12 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 6 & 1 & -1 \\ 23 & 40 & 9 \\ 21 & 37 & 12 \end{pmatrix} \]

### Step 4: Calculate \( \text{tr}(A^2) \)
Now, we can find the trace of \( A^2 \):
\[ \text{tr}(A^2) = 6 + 40 + 12 = 58 \]

### Step 5: Calculate \( \text{tr}(A \cdot \text{adj}(\text{adj}(A))) \)
Now substituting back into our expression:
\[ \text{tr}(A \cdot \text{adj}(\text{adj}(A))) = \det(A) \cdot \text{tr}(A^2) = -28 \cdot 58 \]
Calculating this gives:
\[ -28 \cdot 58 = -1624 \]

### Final Answer
Thus, \( \text{tr}(A \cdot \text{adj}(\text{adj}(A))) = -1624 \).