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(A) Given $(A+B)(A-B) = A^2 - B^2$. Expanding the left side, we get $A^2 - AB + BA - B^2 = A^2 - B^2$. This implies $-AB + BA = 0$, or $AB = BA$. Calc

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(A) Given $(A+B)(A-B) = A^2 - B^2$. Expanding the left side, we get $A^2 - AB + BA - B^2 = A^2 - B^2$. This implies $-AB + BA = 0$, or $AB = BA$. Calculating $AB$: $\begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} \begin{bmatrix} x & y \ 1 & 2 \end{bmatrix} = \begin{bmatrix} x+2 & y+4 \ 2x+1 & 2y+2 \end{bmatrix}$. Calculating $BA$: $\begin{bmatrix} x & y \ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} = \begin{bmatrix} x+2y & 2x+y \ 5 & 4 \end{bmatrix}$. Equating the two matrices: $2x+1 = 5 \Rightarrow 2x = 4 \Rightarrow x = 2$. $2y+2 = 4 \Rightarrow 2y = 2 \Rightarrow y = 1$. Given $C = \begin{bmatrix} x & 2 \ 1 & y \end{bmatrix}$, the trace of $C$ is the sum of its diagonal elements: $\operatorname{Trace}(C) = x + y = 2 + 1 = 3$.

$A=\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} x & y \\ 1 & 2 \end{bmatrix}$ are two matrices such that $(A+B)(A-B)=A^2-B^2$. If $C=\begin{bmatrix} x & 2 \\ 1 & y \end{bmatrix}$, then $\operatorname{Trace}(C)=$

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