Solve the system {x-2y+3z= 4 2x+y-4z= 3 -3x+4y-z= -2

Question

Solve the system

[latex]\left\{\begin{array}{rr}x-2 y+3 z= & 4 \2 x+y-4 z= & 3 \-3 x+4 y-z= & -2\end{array} ight.[/latex]

Accepted Answer

[latex]\left\{\begin{aligned}x-2 y+3 z & =4 \5 y-10 z & =-5 \quad \ -3 x+4 y-z & =-2\end{aligned}\quad \begin{aligned} ext { add }-2 ext { times the first equation }\ ext {to the second equation}\end{aligned} ight.[/latex]

[latex]\left\{\begin{aligned}& x-2 y+3 z= 4 \& 5 y-10 z=-5 \&-2 y+8 z=10\end{aligned}\quad \begin{aligned} ext { add } 3 ext { times the first equation }\ ext {to the third equation}\end{aligned} ight.[/latex]

[latex]\left\{\begin{aligned}x-2 y+3 z= & 4 \y-2 z= & -1 \y-4 z= & -5\end{aligned}\quad \quad \begin{aligned} ext { multiply the second equation by } \frac{1}{5}\ ext { and the third equation by }-\frac{1}{2}\end{aligned} ight.[/latex]

[latex]\left\{\begin{aligned}x-2 y+3 z & =4 \y-2 z & =-1 \-2 z & =-4\end{aligned}\quad \begin{aligned} ext {add -1 times the second equation}\ ext { to the third equation}\end{aligned} ight.[/latex]

[latex]\left\{\begin{aligned}x-2 y+3 z & =4 \y-2 z & =-1 \quad ext { multiply the third equation by }-\frac{1}{2} \z & =2\end{aligned} ight.[/latex]

The solutions of the last system are easy to find by back substitution. From the third equation, we see that z=2. Substituting 2 for z in the second equation, y-2 z=-1, we get y=3. Finally, we find the x-value by substituting y=3 and z=2 in the first equation, x-2y+3z=4, obtaining x=4. Thus, there is one solution, (4, 3, 2).

Question 8.5.1: Solve the system {x-2y+3z= 4 2x+y-4z= 3 -3x+4y-z= -2...

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