Question

Show that the equations $x+y+z=6,x+2y+3z=14,x+4y+7z=30$ are consistent and solve them.

Answer 1

$x+y+z=6,x+2y+3z=14$

$x+4y+7z=30$

$AX=D$

$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 4& 7\end{array}\right]$ $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

Consider Aqument matrix

$AD=\left[\begin{array}{cccc}1& 1& 1& 6\\ 1& 2& 3& 14\\ 1& 4& 7& 30\end{array}\right]$

$R_2:R_2-R_1;R_3:R_3-R_1$

$AD=\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 2& 8\\ 0& 3& 6& 24\end{array}\right]$

$R_3:R_3-3R_2$

$AD=\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 2& 8\\ 0& 0& 0& 0\end{array}\right]$

$\therefore$ Rank of $AD=2$ &

Rank of $A=2$

$\therefore$ It is consistent

$x+y+z=6;y+z=8$

$x+8=6$ $y=k$

$x=-2$ $z=8-k$

$\therefore$ The system is consistent

but has infinite many solutions

$\therefore x=-2,y=k,z=8-k$

$K\in R$ is solution set.