Question:
If \( A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix} \) and \( 2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix} \), then \( \text{Tr}[A] - \text{Tr}[B] \) equals:
If \( A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix} \) and \( 2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix} \), then \( \text{Tr}[A] - \text{Tr}[B] \) equals:
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Use matrix operations systematically: elimination or substitution helps simplify and extract variables. Don't forget that the trace is the sum of the diagonal elements.
Updated On: May 13, 2025
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The Correct Option is B
Solution and Explanation
Step 1: Use matrix equations.
Given: \[ A + 2B = M_1, \quad 2A - B = M_2 \] Multiply equation (1) by 2: \[ 2A + 4B = 2M_1 \] Subtract equation (2): \[ (2A + 4B) - (2A - B) = 2M_1 - M_2 \Rightarrow 5B = 2M_1 - M_2 \Rightarrow B = \frac{1}{5}(2M_1 - M_2) \] Substitute back: \[ A = M_1 - 2B \] Now compute \( \text{Tr}[A] \) and \( \text{Tr}[B] \) from the resulting matrices. \[ \text{Tr}[A] - \text{Tr}[B] = 2 \]
Given: \[ A + 2B = M_1, \quad 2A - B = M_2 \] Multiply equation (1) by 2: \[ 2A + 4B = 2M_1 \] Subtract equation (2): \[ (2A + 4B) - (2A - B) = 2M_1 - M_2 \Rightarrow 5B = 2M_1 - M_2 \Rightarrow B = \frac{1}{5}(2M_1 - M_2) \] Substitute back: \[ A = M_1 - 2B \] Now compute \( \text{Tr}[A] \) and \( \text{Tr}[B] \) from the resulting matrices. \[ \text{Tr}[A] - \text{Tr}[B] = 2 \]
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