91Ó°ÊÓ

The three given equations are:

First, create an augmented version of this system.

The first equation provides us the first row, the second equation gives us the second row, and the third equation gives us the third row in the augmented matrix. The equal signs are changed by a vertical line.

The Argumented matrix is

$\left|\begin{array}{cccc}1& 2& 1& 4\\ 1& 1& -2& 3\\ -2& -3& 1& -7\end{array}\right|$

Consider the argumented matrix:

$$\begin{gathered} \left|\begin{array}{cccc}1& 2& 1& 4\\ 1& 1& -2& 3\\ -2& -3& 1& -7\end{array}\right| \\ \mathrm{Apply}\mathrm{row}\mathrm{operations} \\ \left|\begin{array}{cccc}1& 2& 1& 4\\ 1& 1& -2& 3\\ -2& -3& 1& -7\end{array}\right|\stackrel{{\mathrm{R}}_2→{\mathrm{R}}_1-{\mathrm{R}}_2}{→}\left|\begin{array}{cccc}1& 2& 1& 4\\ 0& 1& 3& 1\\ -2& -3& 1& -7\end{array}\right| \\ \stackrel{{\mathrm{R}}_2→{\mathrm{R}}_1-{\mathrm{R}}_2}{→}\left|\begin{array}{cccc}1& 2& 1& 4\\ 0& 1& 3& 1\\ -2& -3& 1& -7\end{array}\right| \\ \stackrel{{\mathrm{R}}_2→{\mathrm{R}}_1-{\mathrm{R}}_2}{→}\left|\begin{array}{cccc}1& 2& 1& 4\\ 0& 1& 3& 1\\ 0& 1& 3& 1\end{array}\right| \\ \stackrel{{\mathrm{R}}_3→{\mathrm{R}}_2-{\mathrm{R}}_1}{→}\left|\begin{array}{cccc}1& 2& 1& 4\\ 0& 1& 3& 1\\ 0& 0& 0& 0\end{array}\right| \end{gathered}$$

The non-echelon form is

$\left|\begin{array}{cccc}1& 2& 1& 4\\ 0& 1& 3& 1\\ 0& 0& 0& 0\end{array}\right|$

The corresponding equations are:

$$\begin{gathered} x+2y+z=4 \\ y+3z=1 \\ 0=0 \end{gathered}$$

Because the sentence 0=0 is true.

This, like when we solved through substitution, indicates that we are dealing with a dependent system. There are an unlimited number of options.

In the second equation, solve for y in terms of z.

$$\begin{gathered} y+3z=1 \\ y=1-3z \\ \mathrm{Substitute}y=1-3z\mathrm{in}\mathrm{first}\mathrm{equation} \\ x+2y+z=4 \\ x+2y+z=4 \\ x+2(1-3z)+z=4 \\ x+2-6z+z=4 \\ x+2-5z=4 \\ x=5z+4-2 \\ x=5z+2 \end{gathered}$$

Hence ,The solution for the system of linear equations is (5z+2,1-3z,z)

here z is any real number.