computerages Posted February 26, 2008 Posted February 26, 2008 Hello, I was just curious why cramer's rule does not work for infinite solutions. My teacher said that in the class today, but he did not go into details... Can anyone please exaplin that to me... Thanks
NeonBlack Posted February 26, 2008 Posted February 26, 2008 Because if you have an infinite number of solutions, the determinant of the coefficient matrix (the denominator) will be zero. And as you know, division by zero doesn't work. If you want, I'll show an example as soon as I figure out how to write matrices in latex.
ecoli Posted February 26, 2008 Posted February 26, 2008 use \begin{matrix} and \end{matrix} [math] \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} [/math]
timo Posted February 26, 2008 Posted February 26, 2008 From \left[ \begin{array}{ccc} a & b & c \\ d & e & 5 \\ 3 & g & h \end{array} \right] you get [math] \left[ \begin{array}{ccc} a & b & c \\ d & e & 5 \\ 3 & g & h \end{array} \right] [/math]
computerages Posted February 26, 2008 Author Posted February 26, 2008 NeonBlack, it would be great if you can show an example thanks
timo Posted February 26, 2008 Posted February 26, 2008 I think I understand what you meant by "infinite solutions" now, so here goes an example: Equation system: equation 1: 1*x + 1*y = 0 equation 2: 1*x + 1*y = 0 Of course, the system of equations is solved for any x=-y, meaning there is an infinite number of solutions (x=-y=1, x=-y=2, ...). Applying Cramer's rule, you get [math] x = \frac{\text{det} \left[ \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right]}{\text{det} \left[ \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \right]} = 0/0[/math].
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