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Let A(x)=[(0,x-2,x-3),(x+2,0,x-5),(x+3,x...

Let `A(x)=[(0,x-2,x-3),(x+2,0,x-5),(x+3,x+5,0)]`,
then the matrix `A(0)(A(0))^(T)` is a

A

null matrix

B

symmetric matrix

C

skew symmetric matrix

D

non singular matrix

Text Solution

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The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Find A(0) We need to substitute \( x = 0 \) into the matrix \( A(x) \). Given: \[ A(x) = \begin{pmatrix} 0 & x-2 & x-3 \\ x+2 & 0 & x-5 \\ x+3 & x+5 & 0 \end{pmatrix} \] Substituting \( x = 0 \): \[ A(0) = \begin{pmatrix} 0 & 0-2 & 0-3 \\ 0+2 & 0 & 0-5 \\ 0+3 & 0+5 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -2 & -3 \\ 2 & 0 & -5 \\ 3 & 5 & 0 \end{pmatrix} \] ### Step 2: Find A(0) Transpose Now, we need to find the transpose of \( A(0) \). The transpose of a matrix is obtained by swapping rows and columns: \[ A(0)^T = \begin{pmatrix} 0 & 2 & 3 \\ -2 & 0 & 5 \\ -3 & -5 & 0 \end{pmatrix} \] ### Step 3: Calculate A(0) * A(0)^T Next, we will multiply \( A(0) \) by \( A(0)^T \). \[ A(0) = \begin{pmatrix} 0 & -2 & -3 \\ 2 & 0 & -5 \\ 3 & 5 & 0 \end{pmatrix}, \quad A(0)^T = \begin{pmatrix} 0 & 2 & 3 \\ -2 & 0 & 5 \\ -3 & -5 & 0 \end{pmatrix} \] We will perform the multiplication: 1. **First row of \( A(0) \) with each column of \( A(0)^T \)**: - \( (0 \cdot 0) + (-2 \cdot -2) + (-3 \cdot -3) = 0 + 4 + 9 = 13 \) - \( (0 \cdot 2) + (-2 \cdot 0) + (-3 \cdot 5) = 0 + 0 - 15 = -15 \) - \( (0 \cdot 3) + (-2 \cdot 5) + (-3 \cdot 0) = 0 - 10 + 0 = -10 \) 2. **Second row of \( A(0) \) with each column of \( A(0)^T \)**: - \( (2 \cdot 0) + (0 \cdot -2) + (-5 \cdot -3) = 0 + 0 + 15 = 15 \) - \( (2 \cdot 2) + (0 \cdot 0) + (-5 \cdot 5) = 4 + 0 - 25 = -21 \) - \( (2 \cdot 3) + (0 \cdot 5) + (-5 \cdot 0) = 6 + 0 + 0 = 6 \) 3. **Third row of \( A(0) \) with each column of \( A(0)^T \)**: - \( (3 \cdot 0) + (5 \cdot -2) + (0 \cdot -3) = 0 - 10 + 0 = -10 \) - \( (3 \cdot 2) + (5 \cdot 0) + (0 \cdot 5) = 6 + 0 + 0 = 6 \) - \( (3 \cdot 3) + (5 \cdot 5) + (0 \cdot 0) = 9 + 25 + 0 = 34 \) Combining these results, we get: \[ A(0) \cdot A(0)^T = \begin{pmatrix} 13 & -15 & -10 \\ 15 & -21 & 6 \\ -10 & 6 & 34 \end{pmatrix} \] ### Conclusion The resulting matrix \( A(0) \cdot A(0)^T \) is: \[ \begin{pmatrix} 13 & -15 & -10 \\ 15 & -21 & 6 \\ -10 & 6 & 34 \end{pmatrix} \]
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