Understanding the Problem and Key Concepts

This problem asks us to find the value of $Tr(A) - Tr(B)$ given two matrix equations involving matrices $A$ and $B$. The notation $Tr(X)$ denotes the trace of matrix $X$, which is the sum of its diagonal elements.

Before we dive into the solution, let's recall some essential definitions and properties related to matrices and their traces:

1. Matrix Addition/Subtraction: To add or subtract matrices, they must have the same dimensions. The operation is performed element-wise. If $P = [p_{ij}]$ and $Q = [q_{ij}]$ are $m \times n$ matrices, then $P \pm Q = [p_{ij} \pm q_{ij}]$.
2. Scalar Multiplication of a Matrix: To multiply a matrix by a scalar (a number), multiply every element of the matrix by that scalar. If $P = [p_{ij}]$ is an $m \times n$ matrix and $k$ is a scalar, then $kP = [kp_{ij}]$.
3. Trace of a Matrix (Tr(A)): For a square matrix $A$ (i.e., number of rows equals number of columns), its trace is the sum of the elements on its main diagonal. If $A = [a_{ij}]$ is an $n \times n$ matrix, then $Tr(A) = \sum_{i=1}^{n} a_{ii} = a_{11} + a_{22} + \dots + a_{nn}$.
4. Properties of Trace (Linearity): These properties are extremely useful for simplifying calculations involving traces:
   • $Tr(kA) = k \cdot Tr(A)$, where $k$ is a scalar. (The trace of a scalar multiple of a matrix is the scalar multiple of its trace.)
   • $Tr(A+B) = Tr(A) + Tr(B)$. (The trace of a sum of matrices is the sum of their traces.)
   • $Tr(A-B) = Tr(A) - Tr(B)$. (The trace of a difference of matrices is the difference of their traces.) These linearity properties allow us to treat the trace operator somewhat like a linear function, which is key to an efficient solution method.

Given Information

We are provided with two matrix equations: $A + 2B = \left[ {\begin{matrix} 1 & 2 & 0 \\ 6 & { - 3} & 3 \\ { - 5} & 3 & 1 \\ \end{matrix} } \right] \quad \dots(1)$ $2A - B = \left[ {\begin{matrix} 2 & { - 1} & 5 \\ 2 & { - 1} & 6 \\ 0 & 1 & 2 \\ \end{matrix} } \right] \quad \dots(2)$ Let's denote the right-hand side matrices as $M_1$ and $M_2$ respectively for convenience: $M_1 = \left[ {\begin{matrix} 1 & 2 & 0 \\ 6 & { - 3} & 3 \\ { - 5} & 3 & 1 \\ \end{matrix} } \right]$ $M_2 = \left[ {\begin{matrix} 2 & { - 1} & 5 \\ 2 & { - 1} & 6 \\ 0 & 1 & 2 \\ \end{matrix} } \right]$

Our objective is to find the value of $Tr(A) - Tr(B)$. We will demonstrate two methods to solve this problem.

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Method 1: Explicitly Finding Matrices A and B

This method involves solving the system of matrix equations for $A$ and $B$ individually, then calculating their traces, and finally finding the difference. This approach is straightforward but can be more calculation-intensive.

Step 1: Solve for Matrix A

To find $A$, we need to eliminate $B$ from the system of equations. We can treat these matrix equations much like a system of linear algebraic equations.

1. Manipulate equations to eliminate B: We multiply Equation (2) by 2 to make the coefficient of $B$ equal to $-2B$. This will allow $B$ terms to cancel when we add the modified equation to Equation (1). $2 \times (2A - B) = 2 \times M_2$ $4A - 2B = \left[ {\begin{matrix} 2 \times 2 & 2 \times (-1) & 2 \times 5 \\ 2 \times 2 & 2 \times (-1) & 2 \times 6 \\ 2 \times 0 & 2 \times 1 & 2 \times 2 \\ \end{matrix} } \right] = \left[ {\begin{matrix} 4 & { - 2} & {10} \\ 4 & { - 2} & {12} \\ 0 & 2 & 4 \\ \end{matrix} } \right] \quad \dots(3)$

2. Add Equation (1) and Equation (3): Adding the left-hand sides and the right-hand sides of Equation (1) and Equation (3): $(A + 2B) + (4A - 2B) = M_1 + (2M_2)$ $5A = \left[ {\begin{matrix} 1 & 2 & 0 \\ 6 & { - 3} & 3 \\ { - 5} & 3 & 1 \\ \end{matrix} } \right] + \left[ {\begin{matrix} 4 & { - 2} & {10} \\ 4 & { - 2} & {12} \\ 0 & 2 & 4 \\ \end{matrix} } \right]$ $5A = \left[ {\begin{matrix} 1+4 & 2+(-2) & 0+10 \\ 6+4 & -3+(-2) & 3+12 \\ -5+0 & 3+2 & 1+4 \\ \end{matrix} } \right]$ $5A = \left[ {\begin{matrix} 5 & 0 & {10} \\ {10} & { - 5} & {15} \\ { - 5} & 5 & 5 \\ \end{matrix} } \right]$

3. Solve for A: Divide every element of the matrix by 5 (or multiply by $\frac{1}{5}$): $A = \frac{1}{5} \left[ {\begin{matrix} 5 & 0 & {10} \\ {10} & { - 5} & {15} \\ { - 5} & 5 & 5 \\ \end{matrix} } \right] = \left[ {\begin{matrix} 1 & 0 & 2 \\ 2 & { - 1} & 3 \\ { - 1} & 1 & 1 \\ \end{matrix} } \right]$

4. Calculate Tr(A): The trace of $A$ is the sum of its diagonal elements: $a_{11} + a_{22} + a_{33}$. $Tr(A) = 1 + (-1) + 1 = 1$

Step 2: Solve for Matrix B

Similarly, to find $B$, we need to eliminate $A$.

1. Manipulate equations to eliminate A: We multiply Equation (1) by 2 to make the coefficient of $A$ equal to $2A$. Then, we subtract Equation (2) from this modified equation. $2 \times (A + 2B) = 2 \times M_1$ $2A + 4B = \left[ {\begin{matrix} 2 & 4 & 0 \\ {12} & { - 6} & 6 \\ { - 10} & 6 & 2 \\ \end{matrix} } \right] \quad \dots(4)$ Now, subtract Equation (2) from Equation (4): $(2A + 4B) - (2A - B) = (2M_1) - M_2$ $5B = \left[ {\begin{matrix} 2 & 4 & 0 \\ {12} & { - 6} & 6 \\ { - 10} & 6 & 2 \\ \end{matrix} } \right] - \left[ {\begin{matrix} 2 & { - 1} & 5 \\ 2 & { - 1} & 6 \\ 0 & 1 & 2 \\ \end{matrix} } \right]$ $5B = \left[ {\begin{matrix} 2-2 & 4-(-1) & 0-5 \\ 12-2 & -6-(-1) & 6-6 \\ -10-0 & 6-1 & 2-2 \\ \end{matrix} } \right]$ $5B = \left[ {\begin{matrix} 0 & 5 & { - 5} \\ {10} & { - 5} & 0 \\ { - 10} & 5 & 0 \\ \end{matrix} } \right]$

2. Solve for B: Divide every element of the matrix by 5: $B = \frac{1}{5} \left[ {\begin{matrix} 0 & 5 & { - 5} \\ {10} & { - 5} & 0 \\ { - 10} & 5 & 0 \\ \end{matrix} } \right] = \left[ {\begin{matrix} 0 & 1 & { - 1} \\ 2 & { - 1} & 0 \\ { - 2} & 1 & 0 \\ \end{matrix} } \right]$

3. Calculate Tr(B): The trace of $B$ is the sum of its diagonal elements: $b_{11} + b_{22} + b_{33}$. $Tr(B) = 0 + (-1) + 0 = -1$

Step 3: Calculate Tr(A) - Tr(B)

Now that we have $Tr(A)$ and $Tr(B)$, we can find their difference: $Tr(A) - Tr(B) = 1 - (-1) = 1 + 1 = 2$

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Method 2: Using Trace Properties Directly (More Efficient)

This method leverages the linearity properties of the trace operator, allowing us to find $Tr(A)$ and $Tr(B)$ without explicitly calculating the full matrices $A$ and $B$. This often saves time and reduces the chance of arithmetic errors in large matrix calculations.

Step 1: Apply Trace to Equation (1)

Apply the trace operator to both sides of Equation (1): $Tr(A + 2B) = Tr\left( \left[ {\begin{matrix} 1 & 2 & 0 \\ 6 & { - 3} & 3 \\ { - 5} & 3 & 1 \\ \end{matrix} } \right] \right)$ Using the linearity properties $Tr(A+B) = Tr(A) + Tr(B)$ and $Tr(kA) = k \cdot Tr(A)$: $Tr(A) + 2Tr(B) = (1) + (-3) + (1)$ $Tr(A) + 2Tr(B) = -1 \quad \dots(5)$

Step 2: Apply Trace to Equation (2)

Apply the trace operator to both sides of Equation (2): $Tr(2A - B) = Tr\left( \left[ {\begin{matrix} 2 & { - 1} & 5 \\ 2 & { - 1} & 6 \\ 0 & 1 & 2 \\ \end{matrix} } \right] \right)$ Using the linearity properties: $2Tr(A) - Tr(B) = (2) + (-1) + (2)$ $2Tr(A) - Tr(B) = 3 \quad \dots(6)$

Step 3: Solve the System of Linear Equations for Tr(A) and Tr(B)

Now we have a simple system of two linear algebraic equations with two variables, $Tr(A)$ and $Tr(B)$:

1. $Tr(A) + 2Tr(B) = -1$
2. $2Tr(A) - Tr(B) = 3$

To solve for $Tr(A)$ and $Tr(B)$: Multiply Equation (6) by 2: $2 \times (2Tr(A) - Tr(B)) = 2 \times 3$ $4Tr(A) - 2Tr(B) = 6 \quad \dots(7)$ Add Equation (5) and Equation (7): $(Tr(A) + 2Tr(B)) + (4Tr(A) - 2Tr(B)) = -1 + 6$ $5Tr(A) = 5$ $Tr(A) = 1$ Substitute $Tr(A) = 1$ back into Equation (5): $1 + 2Tr(B) = -1$ $2Tr(B) = -2$ $Tr(B) = -1$

Step 4: Calculate Tr(A) - Tr(B)

Finally, calculate the required difference: $Tr(A) - Tr(B) = 1 - (-1) = 1 + 1 = 2$

Both methods yield the same result.

Conclusion and Key Takeaways

The value of $Tr(A) - Tr(B)$ is $\mathbf{2}$.

• Efficiency: For problems involving traces of matrices that are linear combinations of other matrices, directly applying the trace properties (Method 2) is almost always more efficient and less prone to calculation errors than explicitly finding the matrices first (Method 1). It transforms a complex matrix problem into a simpler system of linear algebraic equations involving scalars.
• Fundamental Properties: A solid understanding of matrix addition, scalar multiplication, and especially the linearity properties of the trace operator is crucial for solving such problems efficiently.
• Common Mistake: A common error is to miscalculate matrix sums/differences or scalar multiplications. Double-check all element-wise operations. When using Method 2, ensure the trace of the constant matrices ($M_1$ and $M_2$) is calculated correctly.

The final answer is $\boxed{2}$.