91Ó°ÊÓ

The given system of equation are

$\left\{\begin{array}{l}x-y+z=-4\\ 2x-3y+4z=-15\\ 5x+y-2z=12\end{array}\right.$

The augmented matrix of the system is:

$\left[\begin{array}{cc}\begin{array}{ccc}1& -1& 1\\ 2& -3& 4\\ 5& 1& -2\end{array}& \begin{array}{c}-4\\ -15\\ 12\end{array}\end{array}\right]$

Perform the row operations role="math" localid="1647524178543" $R_2=r_2-r_1,R_3=r_3-5r_1$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& -1& 1\\ 0& -2& 3\\ 0& 6& -7\end{array}& \begin{array}{c}-4\\ -11\\ 32\end{array}\end{array}\right]$

Perform the row operations $R_3=r_3+2r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}1& -1& 1\\ 0& -2& 3\\ 0& 0& 2\end{array}& \begin{array}{c}-4\\ -11\\ -1\end{array}\end{array}\right]$

Use the obtained matrix to write the system of equations.

$$\begin{gathered} x-y+z=-4 \\ -2y+3z=-11 \\ 2z=-1 \end{gathered}$$

Solve the equations to find the solution set.

$$\begin{gathered} 2z=-1 \\ z=-\frac{1}{2} \end{gathered}$$

Substitute $z=-\frac{1}{2}$into second equation.

$$\begin{gathered} -2y+3z=-11 \\ -2y+3\left(-\frac{1}{2}\right)=-11 \\ -2y-\frac{3}{2}=-11 \\ -2y=-11+\frac{3}{2} \\ y=\frac{11}{2}-\frac{3}{4} \\ y=\frac{19}{4} \end{gathered}$$

Substitute $y=\frac{19}{4}\mathrm{and}z=-\frac{1}{2}$into the first equation.

$$\begin{gathered} x-y+z=-4 \\ x-\frac{19}{4}-\frac{1}{2}=-4 \\ x-\frac{17}{4}=-4 \\ x=-4+\frac{17}{4} \\ x=\frac{1}{4} \end{gathered}$$

Therefore, the solution of the system is$x=\frac{1}{4},y=\frac{19}{4},z=-\frac{1}{2}$or, using ordered triplets,$\left(\frac{1}{4},\frac{19}{4},-\frac{1}{2}\right)$.

The solution of the system is $x=\frac{1}{4},y=\frac{19}{4},z=-\frac{1}{2}$or, using ordered triplets,$\left(\frac{1}{4},\frac{19}{4},-\frac{1}{2}\right)$.