1
$\begingroup$

Say I have three equations:

$ x_1A + y_1B + z_1C = 1 $

$ x_2A + y_2B + z_3C = 1 $

$ x_0A + y_0B + z_0C = 1 $

Where $x_1$, $y_1$, $z_1$, ect. are known values.

How do I get each of the three values (A,B,C) from these equations? If someone could take the time to explain their suggested solution that would be great, too.

  • 0
    you must solve this system for $A,B,C$2017-01-15
  • 0
    Yes,that's what I'm asking how to do.2017-01-15
  • 0
    you can use several methods e.g. the Gauss algorithm2017-01-15

2 Answers 2

0

According to Cramer's rule: $$A = \frac{det\left(\begin{matrix}1 & y_1 & z_1\\1 & y_2 & z_2\\1 & y_0 & z_0\end{matrix}\right)}{det\left(\begin{matrix}x_1 & y_1 & z_1\\x_2 & y_2 & z_2\\x_0 & y_0 & z_0\end{matrix}\right)}$$ $$B = \frac{det\left(\begin{matrix}x_1 & 1 & z_1\\x_2 & 1 & z_2\\x_0 & 1 & z_0\end{matrix}\right)}{det\left(\begin{matrix}x_1 & y_1 & z_1\\x_2 & y_2 & z_2\\x_0 & y_0 & z_0\end{matrix}\right)}$$ $$C = \frac{det\left(\begin{matrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_0 & y_0 & 1\end{matrix}\right)}{det\left(\begin{matrix}x_1 & y_1 & z_1\\x_2 & y_2 & z_2\\x_0 & y_0 & z_0\end{matrix}\right)}$$

The Cramer's rule works if ${det\left(\begin{matrix}x_1 & y_1 & z_1\\x_2 & y_2 & z_2\\x_0 & y_0 & z_0\end{matrix}\right)} \ne 0$. Otherwise, the system is linearly dependent and either has no solutions, or infinite number of them.

0

For generality we will assume that the RHS are $t_0,t_1,t_2$ instead of $(1,1,1)$.

Express $A$ from the three equations.

$$A=\frac{t_0-y_0B-z_0C}{x_0}=\frac{t_1-y_1B-z_1C}{x_1}=\frac{t_2-y_2B-z_2C}{x_2}.$$

This gives you a system of two equations in $B,C$ which you can rewrite

$$\left(\frac{y_1}{x_1}-\frac{y_0}{x_0}\right)B+\left(\frac{z_1}{x_1}-\frac{z_0}{x_0}\right)C=\frac{t_1}{x_1}-\frac{t_0}{x_0},\\ \left(\frac{y_2}{x_2}-\frac{y_0}{x_0}\right)B+\left(\frac{z_2}{x_2}-\frac{z_0}{x_0}\right)C=\frac{t_2}{x_2}-\frac{t_0}{x_0}. $$

Now express $B$,

$$B=\frac{\left(\dfrac{t_1}{x_1}-\dfrac{t_0}{x_0}\right)-\left(\dfrac{z_1}{x_1}-\dfrac{z_0}{x_0}\right)C}{\dfrac{y_1}{x_1}-\dfrac{y_0}{x_0}}=\frac{\left(\dfrac{t_2}{x_2}-\dfrac{t_0}{x_0}\right)-\left(\dfrac{z_2}{x_2}-\dfrac{z_0}{x_0}\right)C}{\dfrac{y_2}{x_2}-\dfrac{y_0}{x_0}}$$ and you get a linear equation involving only $C$.

$$\left(\dfrac{\dfrac{z_2}{x_2}-\dfrac{z_0}{x_0}}{\dfrac{y_2}{x_2}-\dfrac{y_0}{x_0}}-\dfrac{\dfrac{z_1}{x_1}-\dfrac{z_0}{x_0}}{\dfrac{y_1}{x_1}-\dfrac{y_0}{x_0}}\right)C=\dfrac{\dfrac{t_2}{x_2}-\dfrac{t_0}{x_0}}{\dfrac{y_2}{x_2}-\dfrac{y_0}{x_0}}-\dfrac{\dfrac{t_1}{x_1}-\dfrac{t_0}{x_0}}{\dfrac{y_1}{x_1}-\dfrac{y_0}{x_0}}$$ which is trivial to solve.

This is the essence of the Gaussian elimination method, which evaluates the Cramer fractions in an efficient way.