If A+2 B=begin{bmatrix} 1 & 2 & 0 6 & -3 & 3 | vedclass

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(B) Given equations are:$A+2B = \begin{bmatrix} 1 & 2 & 0 \ 6 & -3 & 3 \ -5 & 3 & 1 \end{bmatrix}$ ... $(i)$$2A-B = \begin{bmatrix} 2 & -1 & 5 \ 2

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$2$

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(B) Given equations are:$A+2B = \begin{bmatrix} 1 & 2 & 0 \ 6 & -3 & 3 \ -5 & 3 & 1 \end{bmatrix}$ ... $(i)$$2A-B = \begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$ ... $(ii)$Multiply equation $(ii)$ by $2$:$4A-2B = \begin{bmatrix} 4 & -2 & 10 \ 4 & -2 & 12 \ 0 & 2 & 4 \end{bmatrix}$ ... $(iii)$Adding $(i)$ and $(iii)$:$5A = \begin{bmatrix} 1+4 & 2-2 & 0+10 \ 6+4 & -3-2 & 3+12 \ -5+0 & 3+2 & 1+4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \ 10 & -5 & 15 \ -5 & 5 & 5 \end{bmatrix}$$A = \begin{bmatrix} 1 & 0 & 2 \ 2 & -1 & 3 \ -1 & 1 & 1 \end{bmatrix}$$\operatorname{Tr}(A) = 1 + (-1) + 1 = 1$From $(i)$,$2B = (A+2B) - A = \begin{bmatrix} 1 & 2 & 0 \ 6 & -3 & 3 \ -5 & 3 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 2 \ 2 & -1 & 3 \ -1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 & -2 \ 4 & -2 & 0 \ -4 & 2 & 0 \end{bmatrix}$$B = \begin{bmatrix} 0 & 1 & -1 \ 2 & -1 & 0 \ -2 & 1 & 0 \end{bmatrix}$$\operatorname{Tr}(B) = 0 + (-1) + 0 = -1$Therefore,$\operatorname{Tr}(A) - \operatorname{Tr}(B) = 1 - (-1) = 2$.

If $A+2 B=\begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}$ and $2 A-B=\begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$,then $\operatorname{Tr}(A)-\operatorname{Tr}(B) = $

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