Question D.6: Solve for x, y, and z: 1x + 0y - 2z=-1, 0x + 3y + 1z=+2, 1x ...
Solve for x, y, and z:
\begin{aligned}1 x+0 y-2 z=-1 \\0 x+3 y+1 z=+2 \\1 x+2 y+3 z=0 \\\hline\end{aligned}
=\frac{[(-1)(3)(3)+(0)(1)(0)+(-2)(2)(2)]-[(0)(3)(-2)+(2)(1)(-1)+(3)(2)(0)]}{[(1)(3)(3)+(0)(1)(1)+(-2)(0)(2)]-[(1)(3)(-2)+(2)(1)(1)+(3)(0)(0)]}.
=\frac{(-9+0-8)-(0-2+0)}{(9+0+0)-(-6+2+0)}.
=\frac{-17+2}{9+4}=-\frac{15}{13}.
=\frac{[(1)(2)(3)+(-1)(1)(1)+(-2)(0)(0)]-[(1)(2)(-2)+(0)(1)(1)+(3)(0)(-1)]}{13}.
=\frac{(6-1+0)-(-4+0+0)}{13}.
=\frac{5+4}{13}=\frac{9}{13}.
=\frac{[(1)(3)(0)+(0)(2)(1)+(-1)(0)(2)]-[(1)(3)(-1)+(2)(2)(1)+(0)(0)(0)]}{13}.
=\frac{(0+0+0)-(-3+4+0)}{13}.
=\frac{0-1}{13}=-\frac{1}{13}.
\text { or from } 0 x+3 y+1 z=+2.
z=2-3 y=2-3\left(\frac{9}{13}\right)=\frac{26}{13}-\frac{27}{13}=-\frac{1}{13}.
Check:
\left. \begin{aligned}&1 x+0 y-2 z=-1 \\&0 x+3 y+1 z=+2 \\&1 x+2 y+3 z=0\end{aligned} \right\} \left. \begin{aligned}&-\frac{15}{13}+0+\frac{2}{13}=-1 \\&0+\frac{27}{13}+\frac{-1}{13}=+2 \\&-\frac{15}{13}+\frac{18}{13}+\frac{-3}{13}=0\end{aligned} \right\} \begin{aligned}&-\frac{13}{13}=-1 \checkmark \\&\frac{26}{13}=+2 \checkmark \\&-\frac{18}{13}+\frac{18}{13}=0 \checkmark\end{aligned}Related Answered Questions
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