# General Solution

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• Oct 26th 2010, 07:36 PM
l flipboi l
General Solution
Hello,

Can someone please help me figure out the general solution for this matrix?

x+y+z = 0
2x+2y+2z = 0
3x+3y+3z = 0

Thanks!
l Flipboi l
• Oct 26th 2010, 07:57 PM
pickslides
Set it up as

$Ax = b \implies x = A^{-1}b$

where A is the coefficient matrix of $x,y,z$
• Oct 26th 2010, 08:02 PM
l flipboi l
Quote:

Originally Posted by pickslides
Set it up as

$Ax = b \implies x = A^{-1}b$

where A is the coefficient matrix of $x,y,z$

so take the inverse of A and multiple by b?

[ 1 1 1 ]^-1 [ 0 ]
[ 2 2 2 ] [ 0 ]
[ 3 3 3 ] [ 0 ]

the books says the answer is:

x = -s-t
y = s
z = t

what i did earlier was reduced it to row echelon form and obtained the matrix

[ 1 1 1 ]
[ 0 0 0 ]
[ 0 0 0 ]

but i don't get how y=s and z=t
• Oct 27th 2010, 05:18 AM
HallsofIvy
Pickslides, the fact that the problem mentions a "general" solution should tell you that this problem does NOT have one unique solution and so the matrix A is NOT invertible!

|flipboi| you did not write the original equations as a matrix and I see no reason to do so. You are making this much more difficult that it should be.

x+y+z = 0
2x+2y+2z = 0
3x+3y+3z= 0

are all the same equation! So you really just have z= -x- y. That is the "general solution". x and y can be any numbers at all and z= -x- y.

If you really intended
$\begin{bmatrix}1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}0 \\0 \\ 0\end{bmatrix}$ then the vector solution is
$\begin{bmatrix} x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ -x-y\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ -1\end{bmatrix}$

Row reducing to
$\begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$
just tells you, again, that all three equations reduce to x+ y+ z= 0 so that z= -x -y.

"but i don't get how y=s and z= t". I solved for z= -x- y so I could use x and y as parameters or just rename them "s" and "t" and write the solution as x= s, y= t, z= -s- t. Apparently in your text, they solve for x= -y- z. You don't "get" y= s and z= t, you assign those names. Just rename "y" as "s" and "z" as "t". Then x= -s- t, y= s, z= t. There are an infinite number of ways to write parametric equations for a "general" solution.