How would I go about calculating a non-basic solution to something like
2w + 3z = 6
2x + 3z = 8
y + 3z = 7
I'm guessing my solution is going to be a general solution with a free parameter?
Hello, ZTM1989!
You are right!How would I go about calculating a non-basic solution to something like:
. . $\displaystyle \begin{array}{cccc} 2w + 3z &=& 6 & [1] \\
2x + 3z &=& 8 & [2] \\ y + 3z &=& 7 & [3] \end{array}$
I'm guessing my solution is going to be a general solution with a free parameter?
$\displaystyle \begin{array}{ccccccc}\text{Solve [1] for }w: & w &=& \dfrac{6-3z}{2} \\ \\[-3mm]
\text{Solve [2] for }x: & x &=& \dfrac{8-3z}{2} \\ \\[-3mm] \text{Solve [3] for }y: & y &=& 7 - 3z \\ \\[-2mm]
\text{and we have:} & z&=& z \end{array}$
On the right, replace $\displaystyle z$ with a parameter $\displaystyle t.$
And we have: .$\displaystyle \begin{Bmatrix}w &=& \dfrac{6-3t}{2} \\ \\[-3mm]x &=& \dfrac{8-3t}{2} \\ \\[-2mm] y &=& 7-3t \\ \\ z &=& t \end{Bmatrix}$