Set it up as
where A is the coefficient matrix of
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
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.