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Mathematics 16 Online
OpenStudy (anonymous):

The following equations have infinitely many solutions. Give the right hand side of the vector form of the general solution, using a parameter such as s or t.

OpenStudy (anonymous):

9x - 9y - 3z = 12 12x - 6y - z = 28 6x - 12y - 7z = 44

OpenStudy (anonymous):

Is that in a Matrix?

OpenStudy (anonymous):

Yeah, I'm assuming so... Of the form \[\left[\begin{matrix}9 & -9 & -3 &12\\ 12 & -6&-1&28\\6&-12&-7&44\end{matrix}\right]\]

OpenStudy (anonymous):

Also, as an example, for the equations x = y + 1, y = z + 1, z = x − 2 one correct answer is [x, y, z] = [0, −1, −2] + t [1, 1, 1]

OpenStudy (anonymous):

Could this be row reduction I think, you need zeros in bottom left hand corner Triangular matrix, so you can solve for z. Gaussian Elimination or Cramers Rule, does this sound right?

OpenStudy (anonymous):

Yeah, I think so... I can get the matrix down to \[\left[\begin{matrix}1 & 0&1/6&10/3 \\ 0&1&1/2 & 2\\0&0&0&0\end{matrix}\right]\]but not sure what the answer would be... Thanks for the help :)

OpenStudy (anonymous):

t3 = 1 t2 + 1/2t3 = 2 1 + 0.5 = 2 t2 = 0.5 t1 + 1/6t3 = 10/3 1+ 1/6 = 10/3 t1 =3(1/6) z = 1 y = 0.5 x = 0.5 Matrix is linearly dependant infinite solutions exist. As soon as you get all zeros across the bottom row linearly dependant. I think this is right

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