My question is lets say you have
x+5y+6z=10
3x+15y-2z=9
5x+12y-9z=5
So in this example I made up, how would you find the reduced row echeolon form and row echelon form for these equations?
My question is lets say you have
x+5y+6z=10
3x+15y-2z=9
5x+12y-9z=5
So in this example I made up, how would you find the reduced row echeolon form and row echelon form for these equations?
First, write it in matrix format using the coefficients.
$\displaystyle \left[\begin{array}{ccc|c}1&5&6&10\\3&15&-2&9\\5&12&-9&5\end{array}\right]$
The idea is to use elementary row operations to hammer it into the form:
$\displaystyle \left[\begin{array}{ccc|c}1&0&0&a\\0&1&0&b\\0&0&1&c\end{ array}\right]$
Your solutions will be a,b, and c.
I feel about Gaussian eleimination the way Plato feels about partial fraction decompositions. With the technology that abounds these days, why go through the tedium of reduced row echelon?. We can spend our mathematical time more efficiently. But, you gotta do what you gotta do.
It took my TI about 2 seconds to give me:
$\displaystyle \left[\begin{array}{ccc|c}1&0&0&\frac{557}{260}\\0&1&0&\ frac{81}{260}\\0&0&1&\frac{21}{20}\end{array}\righ t]$
I am going to go ahead and show a rref. This is one of many ways to tackle it. You may have a better method with less steps, but this is the idea:
$\displaystyle \left[\begin{array}{ccc|c}1&5&6&10\\3&15&-2&9\\5&12&-9&5\end{array}\right]$
$\displaystyle -5R_{1}+R_{3}\rightarrow{R_{3}}$:
$\displaystyle \left[\begin{array}{ccc|c}1&5&6&10\\3&15&-2&9\\0&-13&-39&-45\end{array}\right]$
$\displaystyle -3R_{1}+R_{2}\rightarrow{R_{2}}$:
$\displaystyle \left[\begin{array}{ccc|c}1&5&6&10\\0&0&-20&-21\\0&-13&-39&-45\end{array}\right]$
$\displaystyle \frac{5}{13}R_{3}+R_{1}\rightarrow{R_{1}}$:
$\displaystyle \left[\begin{array}{ccc|c}1&0&{-9}&\frac{-95}{13}\\0&0&{-20}&{-21}\\0&{-13}&{-39}&{-45}\end{array}\right]$
$\displaystyle \frac{-1}{39}R_{3}\rightarrow{R_{3}}$:
$\displaystyle \left[\begin{array}{ccc|c}1&0&-9&\frac{-95}{13}\\0&0&-20&-21\\0&\frac{13}{39}&1&\frac{45}{39}\end{array}\rig ht]$
$\displaystyle 20R_{3}+R_{2}\rightarrow{R_{2}}$:
$\displaystyle \left[\begin{array}{ccc|c}1&0&-9&\frac{-95}{13}\\0&\frac{20}{3}&0&\frac{27}{13}\\0&\frac{1 3}{39}&1&\frac{45}{39}\end{array}\right]$
$\displaystyle \frac{-1}{20}R_{2}+R_{3}\rightarrow{R_{3}}$:
$\displaystyle \left[\begin{array}{ccc|c}1&0&-9&\frac{-95}{13}\\0&\frac{20}{3}&0&\frac{27}{13}\\0&0&1&\fr ac{21}{20}\end{array}\right]$
$\displaystyle \frac{3}{20}R_{2}\rightarrow{R_{2}}$:
$\displaystyle \left[\begin{array}{ccc|c}1&0&-9&\frac{-95}{13}\\0&1&0&\frac{81}{260}\\0&0&1&\frac{21}{20} \end{array}\right]$
$\displaystyle 9R_{3}+R_{1}\rightarrow{R_{1}}$:
$\displaystyle \left[\begin{array}{ccc|c}1&0&0&\boxed{\frac{557}{260}}\ \0&1&0&\boxed{\frac{81}{260}}\\0&0&1&\boxed{\frac{ 21}{20}}\end{array}\right]$
by the way, the final form galactus put the matrix in is called the reduced row echelon form. it is the form in which the first non-zero digit in each row is 1 and also, all other entries in that 1's column is zero. the row echelon form does not have this last condition. we require the first non-zero digit in any row to be 1 but we do not require that all other entries in the column of the leading 1 be zero.
example:
reduced row echelon form:
$\displaystyle \left[\begin{array}{ccc|c}1&0&0&\frac{557}{260}\\0&1&0&\ frac{81}{260}\\0&0&1&\frac{21}{20}\end{array}\righ t]$
row echelon form:
$\displaystyle \left[\begin{array}{ccc|c}1&{\color{red}2}&0&\frac{557}{ 260}\\0&1&{\color{red}4}&\frac{81}{260}\\0&0&1&\fr ac{21}{20}\end{array}\right]$
the 2 and the 4 in the above matrix prevents it from being in reduced row echelon form.
in general, it is easier to solve a problem by bringing the matrix in reduced row echelon form if possible, otherwise, you will need to back substitute to get your solutions if it is only in row echelon form