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}