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 \( \text{tr}(A) - \text{tr}(B) = \)

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Using the linearity of the trace operator avoids the need to solve for the full matrices \( A \) and \( B \) element by element.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

The trace of a matrix, denoted as \( \text{tr}(X) \), is the sum of its diagonal elements. The trace is a linear operator, meaning \( \text{tr}(A + B) = \text{tr}(A) + \text{tr}(B) \) and \( \text{tr}(kA) = k \cdot \text{tr}(A) \).
Step 2: Detailed Explanation:

Let \( t_A = \text{tr}(A) \) and \( t_B = \text{tr}(B) \). From the first equation: \[ \text{tr}(A + 2B) = \text{tr}\begin{bmatrix} 1 & 2 & 0
6 & -3 & 3
-5 & 3 & 1 \end{bmatrix} \] \[ t_A + 2t_B = 1 + (-3) + 1 = -1 \quad \dots (1) \] From the second equation: \[ \text{tr}(2A - B) = \text{tr}\begin{bmatrix} 2 & -1 & 5
2 & -1 & 6
0 & 1 & 2 \end{bmatrix} \] \[ 2t_A - t_B = 2 + (-1) + 2 = 3 \quad \dots (2) \] Now we solve the system of linear equations for \( t_A \) and \( t_B \): From (2), \( t_B = 2t_A - 3 \). Substitute this into (1): \[ t_A + 2(2t_A - 3) = -1 \] \[ t_A + 4t_A - 6 = -1 \] \[ 5t_A = 5 \implies t_A = 1 \] Now find \( t_B \): \[ t_B = 2(1) - 3 = -1 \] We need the value of \( \text{tr}(A) - \text{tr}(B) \): \[ t_A - t_B = 1 - (-1) = 2 \]
Step 4: Final Answer:

The value is 2.