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Mathematics 19 Online
OpenStudy (anonymous):

form a solution to the given system for any choice of the free parameter(s) −2x1+3x2+2x3=6

OpenStudy (anonymous):

Let \(x_1\) and \(x_2\) be your free parameters.

OpenStudy (anonymous):

First let \(x_1=0, x_2=1\). Then \[ -2(0)+3(1)+2x_3=6 \implies 3+2x_3=6\implies x_3=\frac 3 2= 1.5 \]So our first vector is \[ \mathbf v_0=\begin{bmatrix}0 \\ 1 \\ 1.5\end{bmatrix} \]

OpenStudy (anonymous):

Then let \(x_1=1, x_2=0\) and find the next vector. Does that help so far? @paki123

OpenStudy (anonymous):

Kind of. I'm not really sure what to do after. Wy did you make x1 = 0, and x2 =1?

OpenStudy (anonymous):

They can be anything you want them to be, but they need to be linearly independent.

OpenStudy (anonymous):

0,1 and 1,0 are orthogonal, so they are clearly linearly independent.

OpenStudy (anonymous):

Could i make them equal s1, and s2? Since the answe will have to be in that format?

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