d
(d) Let $A = \left[ {\begin{array}{*{20}{c}}1&0&1\\{ - 1}&1&0\\0&{ - 1}&1\end{array}} \right]\, \Rightarrow \,{A^{ - 1}}\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&{\frac{{ - 1}}{2}}&{ - 1}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{{ - 1}}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\end{array}} \right]$
$\therefore $ $AX = B \Rightarrow X = {A^{ - 1}}B\,\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&{\frac{{ - 1}}{2}}&{ - 1}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{{ - 1}}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\end{array}} \right]\,\,\left[ \begin{array}{l}1\\1\\2\end{array} \right] = \left[ \begin{array}{l} - 1\\{\rm{ }}0\\{\rm{ }}2\end{array} \right]$.
Aliter : $\left[ {\begin{array}{*{20}{c}}1&0&1\\{ - 1}&1&0\\0&{ - 1}&1\end{array}} \right]\,\,\,\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}1\\1\\2\end{array} \right]$
$ \Rightarrow $ $\left[ \begin{array}{l}x + 0y + z\\ - x + y + 0z\\0x - y + z\end{array} \right] = \left[ \begin{array}{l}1\\1\\2\end{array} \right]$ ==> $\begin{array}{l}x + z = 1\\ - x + y = 1\\z - y = 2\end{array}$
$ \Rightarrow $ $(x,\,y,\,z) = ( - 1,\,0,\,2)$.