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Question Number 204056 by hardmath last updated on 05/Feb/24

$$\{\begin{array}{l}\mathrm{x}+2\mathrm{y}+\mathrm{z}=8\\ 3\mathrm{x}+2\mathrm{y}+\mathrm{z}=10\\ 4\mathrm{x}+3\mathrm{y}-2\mathrm{z}=4\end{array}$$$$\mathrm{Solve}\mathrm{with}\mathrm{the}\mathrm{help}\mathrm{of}\mathrm{matrix}$$

Answered by deleteduser1 last updated on 05/Feb/24

$$\left[\begin{array}{ccc}1& 2& 1\\ 3& 2& 1\\ 4& 3& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}8\\ 10\\ 4\end{array}\right]\underset{R_3-4R_1}{\stackrel{R_2-3R_1}{\Rightarrow }}\left[\begin{array}{cccc}1& 2& 1& 8\\ 0& -4& -2& -14\\ 0& -5& -6& -28\end{array}\right]$$$$\underset{4R_3-5R_2}{\stackrel{2R_1+R_2}{\Rightarrow }}\left[\begin{array}{cccc}2& 0& 0& 2\\ 0& -4& -2& -14\\ 0& 0& -14& -42\end{array}\right]\stackrel{-7R_2+R_3}{\Rightarrow }\left[\begin{array}{ccccc}2& 0& 0& \mid & 2\\ 0& 28& 0& \mid & 56\\ 0& 0& -14& \mid & -42\end{array}\right]$$$$\Rightarrow 2x+0y+0z=2\Rightarrow x=1;0x+28y+0z=56\Rightarrow y=2$$$$0x+0y-14z=-42\Rightarrow z=3\Rightarrow (x,y,z)=(1,2,3).$$

Answered by MikeH last updated on 05/Feb/24

$$\left(\begin{array}{ccc}1& 2& 1\\ 3& 2& 1\\ 4& 3& -2\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}8\\ 10\\ 4\end{array}\right)$$$$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)={\left(\begin{array}{ccc}1& 2& 1\\ 3& 2& 1\\ 4& 3& -2\end{array}\right)}^{-1}\left(\begin{array}{c}8\\ 10\\ 4\end{array}\right)$$$$\Rightarrow \left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}-\frac{1}{2}& \frac{1}{2}& 0\\ \frac{5}{7}& -\frac{3}{7}& \frac{1}{7}\\ \frac{1}{14}& \frac{5}{14}& -\frac{2}{7}\end{array}\right)\left(\begin{array}{c}8\\ 10\\ 4\end{array}\right)=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$$

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