I need help solving this problem

[Mougel]

Solve: 2x-y+3z=8
x-6y-z=0
-6x+3y-9z=24

Every time I try all the varibles cancel out.[/list]

 

[soroban]

Hello, Mougel!

Solve: \(\displaystyle \:\begin{array}{ccc}2x\,-\,y\,+\,3z &\,=\, & 8 \\
x\,-\,6y\,-\,z & \,=\, & 0 \\-6x\,+\,3y\,-\.9z & \,=\, &24\end{array}\)

Every time I try, all the varibles cancel out. . Right!

Your work is correct . . . The determinant of the system is zero.

\(\displaystyle \;\;\;\;\begin{vmatrix}2 & -1 & 3 \\ 1 & -6 & -1 \\ -6 & 3 & -9\end{vmatrix} \:=\:0\)

Therefore, the system of equations has no solution.

 

[Denis]

2x - y + 3z = 8 [1]
-6x + 3y - 9z = 24 [2]

No need to use determinants and the likes; equation [2] is really:
-3(2x - y + 3z) = 24
2x - y + 3z = -8

Now look at equation [1] !

 

[stapel]

...The determinant of the system is zero.... Therefore, the system of equations has no solution.

Actually, a zero determinant, as far as I know, only means that the system has no unique solution. The system might actually be dependent. For instance, the following system has a solution:

. . . . .2x + 2y = 4
. . . . .4x + 4x = 8

Clearly the solution is the line y = -x + 2, or (x, y) = (s, 2 - s). But the determinant is 8 - 8 = 0.

A zero determinant, in this context, indicates the need to solve the system of equations without using determinants, to discover whether there is no solution (so the system is "indeterminate") or a non-unique solution (so the system is "dependent"). You cannot, to my knowledge, conclude the solution type from the zero determinant.

Eliz.

P.S. to Mougel: Welcome to FreeMathHelp!