Let A+2 B=left[begin{array}{ccc}1 & 2 & 0 6 | vedclass

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

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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} \quad \dots(1)$$2A - B = \begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix} \quad \dots(2)$Multiply equation $(2)$ by $2$:$4A - 2B = \begin{bmatrix} 4 & -2 & 10 \ 4 & -2 & 12 \ 0 & 2 & 4 \end{bmatrix} \quad \dots(3)$Adding $(1)$ and $(3)$:$5A = \begin{bmatrix} 1+4 & 2-2 & 0+10 \ 6+4 & -3-2 & 3+12 \ -5+0 & 3+2 & 1+4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \ 10 & -5 & 15 \ -5 & 5 & 5 \end{bmatrix}$$A = \begin{bmatrix} 1 & 0 & 2 \ 2 & -1 & 3 \ -1 & 1 & 1 \end{bmatrix}$Now,from $(2)$,$B = 2A - \begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$:$B = \begin{bmatrix} 2 & 0 & 4 \ 4 & -2 & 6 \ -2 & 2 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -1 \ 2 & -1 & 0 \ -2 & 1 & 0 \end{bmatrix}$$\operatorname{Tr}(A) = 1 + (-1) + 1 = 1$$\operatorname{Tr}(B) = 0 + (-1) + 0 = -1$$\operatorname{Tr}(A) - \operatorname{Tr}(B) = 1 - (-1) = 2$

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