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

Question

Solve the system

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

Accepted Answer

We begin with the matrix of the system—that is, with the augmented matrix:

[latex]\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \2 & 1 & -4 & 3 \-3 & 4 & -1 & -2\end{array} ight][/latex]

We next apply elementary row transformations to obtain another (simpler) matrix of an equivalent system of equations. These transformations correspond to the manipulations used for equations in Example 1. We will place appropriate symbols between equivalent matrices.

[latex]\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \2 & 1 & -4 & 3 \-3 & 4 & -1 & -2\end{array} ight] \begin{aligned}& -2 {R}_1+{R}_2 ightarrow {R}_2 \& 3 {R}_1+{R}_3 ightarrow {R}_3\end{aligned}\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \0 & 5 & -10 & -5 \0 & -2 & 8 & 10\end{array} ight] \quad \begin{aligned}& ext { add }-2 R_1 ext { to } R_2 \& ext { add } 3 R_1 ext { to } R_3\end{aligned}[/latex]

[latex]\begin{aligned}& \begin{array}{r}\frac{1}{5} {R}_2 ightarrow {R}_2 \-\frac{1}{2} {R}_3 ightarrow {R}_3\end{array} \left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \0 & 1 & -2 & -1 \0 & 1 & -4 & -5\end{array} ight] \quad \begin{array}{l} ext { multiply } \mathrm{R}_2 ext { by } \frac{1}{5} \ ext { multiply } \mathrm{R}_3 ext { by }-\frac{1}{2}\end{array} \& -{R}_2+{R}_3 ightarrow{R}_3\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \0 & 1 & -2 & -1 \0 & 0 & -2 & -4\end{array} ight] \quad ext { add }-\mathrm{R}_2 ext { to } \mathrm{R}_3 \& -\frac{1}{2} {R}_3 ightarrow {R}_3\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \0 & 1 & -2 & -1 \0 & 0 & 1 & 2\end{array} ight] \quad ext { multiply } \mathrm{R}_3 ext { by }-\frac{1}{2} \&\end{aligned}[/latex]

We use the last matrix to return to the system of equations

[latex]\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \0 & 1 & -2 & -1 \0 & 0 & 1 & 2\end{array} ight]\Longleftrightarrow\left\{\begin{aligned}x-2 y+3 z= & 4 \y-2 z= & -1 \z= & 2\end{aligned} ight.[/latex]

which is equivalent to the original system. The solution x=4, y=3, z=2 may now be found by back substitution, as in Example 1.

Question 8.5.2: Solve the system {x-2y+3z= 4 2x+y-4z=3 -3x+4y-z= -2...

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