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

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- October 26th 2010, 07:36 PMl flipboi lGeneral 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 - October 26th 2010, 07:57 PMpickslides
Set it up as

where A is the coefficient matrix of - October 26th 2010, 08:02 PMl flipboi l
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 - October 27th 2010, 05:18 AMHallsofIvy
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

then the vector solution is

Row reducing to

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.