Show that the system of equation $\displaystyle
x - 3y -8z = -10;
$$\displaystyle
3x+y-4z=0;
$ and $\displaystyle
2x+5y+6z =13
$ is consistent and find the solution .
OK, the determinamt is equal to zero (I should have checked). This means that either there's no solution or an infinite number of solutions. So my suggestion is no good after all.
In fact, since x = 3y + 8z from the first equation, the second and third equations become:
y + 2z = 0 and y + 2z = 13/2.
Therefore the equations are INconsistent and there is no solution to be found.
HI
$\displaystyle x-3y-8z=-10$ ---- 1
$\displaystyle 3x+y-4z=0$ -----2
$\displaystyle 2x+5y+6z=13$ ---- 3
1x2 -3 $\displaystyle y+2z=3\Rightarrow z=\frac{3-y}{2}$ ---4
Substitute 4 into equation 3 .
$\displaystyle 2x+5y+6(\frac{3-y}{2})=13$
$\displaystyle x=2-y$
so (x,y,z)=(2-y , y , (3-y)/2)
(x,y,z)=y(-1,1,-1/2)+(2,0,3/2) where $\displaystyle y\in R$
and if y=1 , (x,y,z)=(1,1,1) which agrees with CB .
Not sure if this is right though .
I'm not sure if this is a right approach to this problem due to teaching myself the material but could you not just use gauss-jordan elimination.
The last row would be
$\displaystyle 0x +0y + 0z = 0$
therefore showing that there is infinite number of solutions