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codysmom99 · Oct 25, 2009, 07:36 PM

How do you solve this system by inverses?

x + y + z = 1
2x + y = -2
3y + z = 2

Perito · Oct 25, 2009, 07:49 PM

I'm not familiar with the term "inverse" for solving simultaneous equations. Are you using Cramer's rule (matrices/determinants) to solve this?

I'm assuming that this is what you mean, but I'm not sure.

Set up a matrix from the coefficients of the three equations

x + y + z = 1
2x + y = -2
3y + z = 2

$\left[ \begin{matrix}1 & 1 & 1 \\ 2 & 1 & 0 \\ 0 & 3 & 1 \\ \end{matrix} \right]$

Calculate the determinate of this matrix (5, if I did it correctly).

$\left| \begin{matrix}1 & 1 & 1 \\ 2 & 1 & 0 \\ 0 & 3 & 1 \\ \end{matrix} \right| = 1 + 6 + 0 - 0 - 2 - 0 = 5$

To solve for X, replace the first column with a vector of the numbers on the right of the equations. Divide the resulting determinant by the determinant calculated above (5)

$\left[ \begin{matrix}1 \\ -2 \\ 2 \\ \end{matrix} \right]$

$x = \frac {\left| \begin{matrix}1 & 1 & 1 \\ -2 & 1 & 0 \\ 2 & 3 & 1 \\ \end{matrix} \right|}{5}$

Replace the middle column to solve for Y

$y = \frac {\left| \begin{matrix}1 & 1 & 1 \\ 2 & -2 & 0 \\ 0 & 2 & 1 \\ \end{matrix} \right|}{5}$

Replace the rightmost column to solve for Z

$z = \frac {\left| \begin{matrix}1 & 1 & 1 \\ 2 & 1 & -2 \\ 0 & 3 & 2 \\ \end{matrix} \right|}{5}$