91Ó°ÊÓ

A system of linear equations is given as-

$\left\{\begin{array}{l}2x-3y+z=19\\ -3x+y-2z=-15\\ x+y+z=0\end{array}\right.$

First, we will convert the given system of linear equations into its argument matrix. After that we will apply row transformation accordingly.

And we will to make zeroes in the matrix at least two in one row and one in other row through which we can find the value of one of the variable. And then we will find the value of other variable by using value of first one.

A system of linear equations is given as-

$\left\{\begin{array}{l}2x-3y+z=19\\ -3x+y-2z=-15\\ x+y+z=0\end{array}\right.$

Correspond argument matrix is -

$\left[\begin{array}{rrrr}2& -3& 1& 19\\ -3& 1& -2& -15\\ 1& 1& 1& 0\end{array}\right]$

Apply row transformations as-

$$\begin{gathered} \left[\begin{array}{rrrc}2& -3& 1& 19\\ -3& 1& -2& -15\\ 1& 1& 1& 0\end{array}\right]\stackrel{R_1↔R_2}{âŸ\P }\left[\begin{array}{rrrr}1& 1& 1& 0\\ -3& 1& -2& -15\\ 2& -3& 1& 19\end{array}\right] \\ \stackrel{3R_1+R_2}{âŸ\P }\left[\begin{array}{rrrr}1& 1& 1& 0\\ 0& 4& 1& -15\\ 2& -3& 1& 19\end{array}\right] \\ \stackrel{-2R_1+R_1}{âŸ\P }\left|\begin{array}{rrrr}1& 1& 1& 0\\ 0& 4& 1& -15\\ 0& -5& -1& 19\end{array}\right| \\ \stackrel{R_2+R_1}{âŸ\P }\left[\begin{array}{cccc}1& 1& 1& 0\\ 0& 4& 1& -15\\ 0& -1& 0& 4\end{array}\right] \\ \stackrel{-R_3}{âŸ\P }\left[\begin{array}{lllc}1& 1& 1& 0\\ 0& 4& 1& -15\\ 0& 1& 0& -4\end{array}\right] \end{gathered}$$

Accordingly, the equations are-

$$\begin{gathered} x+y+z=0 \\ 4y+z=-15 \\ y=-4 \end{gathered}$$

Use substitution,

$$\begin{gathered} 4y+z=-15 \\ 4(-4)+z=-15 \\ -16+z=-15 \\ z=-15+16 \\ z=1 \end{gathered}$$

For the value of $x$,

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

Verify the equation,

$$\begin{gathered} 2x-3y+z=19 \\ 2\left(3\right)-3(-4)+1=19 \\ 6+12+1=19 \\ 19=19 \end{gathered}$$

LHS is equal to RHS