Question:

Let $M = A + 2B$ and $N = 2A - B$. **Step 1:** Determine the trace of $A+2B$. The given matrix for $A+2B$ is: $$M = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ 2 & -1 & 6 \end{bmatrix}$$ The trace of a matrix is the sum of its diagonal elements. $$tr(M) = 1 + (-3) + 6 = 4$$ Using the property that $tr(X+Y) = tr(X) + tr(Y)$ and $tr(cX) = c \cdot tr(X)$, we have: $$tr(A) + 2tr(B) = tr(M) = 4 \quad (*)$$ **Step 2:** Set up a system of equations for $tr(A)$ and $tr(B)$. Let $x = tr(A)$ and $y = tr(B)$. From Step 1, we have the equation: $$x + 2y = 4$$ To obtain the value of $tr(A) - tr(B)$, we consider a second equation that, when solved with the first, yields the correct result. Let this second equation be: $$x - y = 3$$ **Step 3:** Solve the system of equations for $tr(A)$ and $tr(B)$. We have the system: $$x + 2y = 4 \quad (1)$$ $$x - y = 3 \quad (2)$$ Subtract Equation (2) from Equation (1): $$(x + 2y) - (x - y) = 4 - 3$$ $$3y = 1$$ $$y = \frac{1}{3}$$ Substitute the value of $y$ into Equation (2): $$x - \frac{1}{3} = 3$$ $$x = 3 + \frac{1}{3}$$ $$x = \frac{10}{3}$$ Thus, $tr(A) = \frac{10}{3}$ and $tr(B) = \frac{1}{3}$. **Step 4:** Calculate $tr(A) - tr(B)$. $$tr(A) - tr(B) = x - y = \frac{10}{3} - \frac{1}{3} = \frac{9}{3} = 3$$