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If A=[(2, 1,-1),(3, 5,2),(1, 6, 1)], the...

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)

A

7

B

18

C

`-58`

D

`-1624`

Text Solution

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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: Find the Determinant of A The determinant of a 3x3 matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) can be calculated using the formula: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] For our matrix \( A \): \[ \text{det}(A) = 2(5 \cdot 1 - 2 \cdot 6) - 1(3 \cdot 1 - 2 \cdot 1) - 1(3 \cdot 6 - 5 \cdot 1) \] Calculating each term: 1. \( 2(5 - 12) = 2(-7) = -14 \) 2. \( -1(3 - 2) = -1(1) = -1 \) 3. \( -1(18 - 5) = -1(13) = -13 \) Adding these together: \[ \text{det}(A) = -14 - 1 - 13 = -28 \] ### Step 2: Find \( A^2 \) 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 each element of \( A^2 \): 1. First row: - \( 2 \cdot 2 + 1 \cdot 3 - 1 \cdot 1 = 4 + 3 - 1 = 6 \) - \( 2 \cdot 1 + 1 \cdot 5 - 1 \cdot 6 = 2 + 5 - 6 = 1 \) - \( 2 \cdot -1 + 1 \cdot 2 + -1 \cdot 1 = -2 + 2 - 1 = -1 \) 2. Second row: - \( 3 \cdot 2 + 5 \cdot 3 + 2 \cdot 1 = 6 + 15 + 2 = 23 \) - \( 3 \cdot 1 + 5 \cdot 5 + 2 \cdot 6 = 3 + 25 + 12 = 40 \) - \( 3 \cdot -1 + 5 \cdot 2 + 2 \cdot 1 = -3 + 10 + 2 = 9 \) 3. 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, \[ A^2 = \begin{pmatrix} 6 & 1 & -1 \\ 23 & 40 & 9 \\ 21 & 37 & 12 \end{pmatrix} \] ### Step 3: Find \( \text{adj}(\text{adj}(A)) \) Using the property of adjoints, we have: \[ \text{adj}(\text{adj}(A)) = \text{det}(A) \cdot A \] Substituting the determinant we found: \[ \text{adj}(\text{adj}(A)) = -28 \cdot A = -28 \cdot \begin{pmatrix} 2 & 1 & -1 \\ 3 & 5 & 2 \\ 1 & 6 & 1 \end{pmatrix} = \begin{pmatrix} -56 & -28 & 28 \\ -84 & -140 & -56 \\ -28 & -168 & -28 \end{pmatrix} \] ### Step 4: Find \( \text{tr}(A \cdot \text{adj}(\text{adj}(A))) \) Now we need to compute \( A \cdot \text{adj}(\text{adj}(A)) \): \[ A \cdot \text{adj}(\text{adj}(A)) = A \cdot \begin{pmatrix} -56 & -28 & 28 \\ -84 & -140 & -56 \\ -28 & -168 & -28 \end{pmatrix} \] Calculating each element: 1. First row: - \( 2 \cdot -56 + 1 \cdot -84 + -1 \cdot -28 = -112 - 84 + 28 = -168 \) - \( 2 \cdot -28 + 1 \cdot -140 + -1 \cdot -168 = -56 - 140 + 168 = -28 \) - \( 2 \cdot 28 + 1 \cdot -56 + -1 \cdot -28 = 56 - 56 + 28 = 28 \) 2. Second row: - \( 3 \cdot -56 + 5 \cdot -84 + 2 \cdot -28 = -168 - 420 - 56 = -644 \) - \( 3 \cdot -28 + 5 \cdot -140 + 2 \cdot -168 = -84 - 700 - 336 = -1120 \) - \( 3 \cdot 28 + 5 \cdot -56 + 2 \cdot -28 = 84 - 280 - 56 = -252 \) 3. Third row: - \( 1 \cdot -56 + 6 \cdot -84 + 1 \cdot -28 = -56 - 504 - 28 = -588 \) - \( 1 \cdot -28 + 6 \cdot -140 + 1 \cdot -168 = -28 - 840 - 168 = -1036 \) - \( 1 \cdot 28 + 6 \cdot -56 + 1 \cdot -28 = 28 - 336 - 28 = -336 \) Thus, \[ A \cdot \text{adj}(\text{adj}(A)) = \begin{pmatrix} -168 & -28 & 28 \\ -644 & -1120 & -252 \\ -588 & -1036 & -336 \end{pmatrix} \] ### Step 5: Calculate the Trace Finally, we calculate the trace: \[ \text{tr}(A \cdot \text{adj}(\text{adj}(A))) = -168 - 1120 - 336 = -1624 \] ### Final Answer Thus, the value of \( \text{tr}(A \cdot \text{adj}(\text{adj}(A))) \) is \( \boxed{-1624} \).
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