Use back-substitution to solve the triangular system. $$\left\{ \begin{aligned}
x - 3 y - z & = 0 \
2 y - 8 z & = 2 \
\frac { 1 } { 4 } z & = - 1
\end{aligned} \right.$$

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The Answer of Use back-substitution to solve the triangular system. $$\left\{ \begin{aligned}
x - 3 y - z & = 0 \
2 y - 8 z & = 2 \
\frac { 1 } { 4 } z & = - 1
\end{aligned} \right.$$

Use Back-Substitution to Solve the Triangular System {x−3y−z=02y−8z=214z=−1\left\{ \begin{aligned}x - 3 y - z & = 0 \\2 y - 8 z & = 2 \\\frac { 1 } { 4 } z & = - 1\end{aligned} \right.⎩⎨⎧​x−3y−z2y−8z41​z​=0=2=−1​

Use Back-Substitution to Solve the Triangular System $\left\{ \begin{aligned}x - 3 y - z & = 0 \\2 y - 8 z & = 2 \\\frac { 1 } { 4 } z & = - 1\end{aligned} \right.$