If S = begin{bmatrix} 0 & 1 & 1 1 & 0 & 1 1 | vedclass

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(A) First,we find the inverse of $S$. The determinant $|S| = 0(0-1) - 1(0-1) + 1(1-0) = 1 + 1 = 2$.The adjugate of $S$ is $	ext{adj}(S) = \begin{bmat

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$\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}$

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(A) First,we find the inverse of $S$. The determinant $|S| = 0(0-1) - 1(0-1) + 1(1-0) = 1 + 1 = 2$.The adjugate of $S$ is $	ext{adj}(S) = \begin{bmatrix} -1 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & -1 \end{bmatrix}$.Thus,$S^{-1} = \frac{1}{2} \begin{bmatrix} -1 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & -1 \end{bmatrix}$.Next,we compute $SA = \frac{1}{2} \begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} b+c & c-a & b-a \ c-b & c+a & a-b \ b-c & a-c & a+b \end{bmatrix} = \begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix} S$.Alternatively,calculating $SA$ directly gives $\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix} S$.Therefore,$SAS^{-1} = (SA)S^{-1} = \begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix} S S^{-1} = \begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix} I = \begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}$.

If $S = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$ and $A = \frac{1}{2} \begin{bmatrix} b+c & c-a & b-a \\ c-b & c+a & a-b \\ b-c & a-c & a+b \end{bmatrix}$,then $SAS^{-1} =$

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