The solution of the equation left[begin{array | vedclass

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(D) Given the matrix equation: $\left[\begin{array}{rrr}1 & 0 & 1 \ -1 & 1 & 0 \ 0 & -1 & 1\end{array}ight]\left[\begin{array}{l}x \ y \ z\end{a

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$(-1, 0, 2)$

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(D) Given the matrix equation: $\left[\begin{array}{rrr}1 & 0 & 1 \ -1 & 1 & 0 \ 0 & -1 & 1\end{array}ight]\left[\begin{array}{l}x \ y \ z\end{array}ight]=\left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$ Multiplying the matrices,we get the system of linear equations: $x + z = 1$  $(i)$ $-x + y = 1$  $(ii)$ $-y + z = 2$  $(iii)$ Adding $(ii)$ and $(iii)$,we get: $(-x + y) + (-y + z) = 1 + 2$ $-x + z = 3$  $(iv)$ Now,adding $(i)$ and $(iv)$: $(x + z) + (-x + z) = 1 + 3$ $2z = 4 \Rightarrow z = 2$ Substituting $z = 2$ into $(i)$: $x + 2 = 1 \Rightarrow x = -1$ Substituting $x = -1$ into $(ii)$: $-(-1) + y = 1 \Rightarrow 1 + y = 1 \Rightarrow y = 0$ Thus,the solution is $(x, y, z) = (-1, 0, 2)$.

The solution of the equation $\left[\begin{array}{rrr}1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$ is $(x, y, z)=$

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