Let 2A+B = begin{bmatrix} 1 & 0 & 3 -1 & 4 & | vedclass

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(B) The trace of a matrix,$Tr(M)$,is the sum of its diagonal elements. Given $2A + B = \begin{bmatrix} 1 & 0 & 3 \ -1 & 4 & 6 \ 2 & 5 & 2 \end{bmatr

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$5$

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(B) The trace of a matrix,$Tr(M)$,is the sum of its diagonal elements. Given $2A + B = \begin{bmatrix} 1 & 0 & 3 \ -1 & 4 & 6 \ 2 & 5 & 2 \end{bmatrix}$,the trace is $Tr(2A + B) = 1 + 4 + 2 = 7$. Since $Tr(2A + B) = 2Tr(A) + Tr(B)$,we have $2Tr(A) + Tr(B) = 7$ (Equation $1$). Given $A - 2B = \begin{bmatrix} 2 & -1 & 5 \ 0 & 3 & 6 \ 1 & 2 & 1 \end{bmatrix}$,the trace is $Tr(A - 2B) = 2 + 3 + 1 = 6$. Since $Tr(A - 2B) = Tr(A) - 2Tr(B)$,we have $Tr(A) - 2Tr(B) = 6$ (Equation $2$). Multiplying Equation $1$ by $2$: $4Tr(A) + 2Tr(B) = 14$. Adding this to Equation $2$: $(4Tr(A) + 2Tr(B)) + (Tr(A) - 2Tr(B)) = 14 + 6 \Rightarrow 5Tr(A) = 20 \Rightarrow Tr(A) = 4$. Substituting $Tr(A) = 4$ into Equation $1$: $2(4) + Tr(B) = 7 \Rightarrow 8 + Tr(B) = 7 \Rightarrow Tr(B) = -1$. Finally,$Tr(A) - Tr(B) = 4 - (-1) = 4 + 1 = 5$.

Let $2A+B = \begin{bmatrix} 1 & 0 & 3 \\ -1 & 4 & 6 \\ 2 & 5 & 2 \end{bmatrix}$ and $A-2B = \begin{bmatrix} 2 & -1 & 5 \\ 0 & 3 & 6 \\ 1 & 2 & 1 \end{bmatrix}$. Then $Tr(A) - Tr(B)$ has the value equal to (where $Tr(A)$ denotes the trace of matrix $A$).

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Let $A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}$ and $2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$. Then $Tr(A) - Tr(B)$ has the value equal to:

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