Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 27/Solution 2

Since the solution is $x=[0,1,0,1{]}^{T}s+[7,0,1,0{]}^{T}t$, we can rewrite it as $x=\left[\begin{array}{c}7t\\ s\\ t\\ s\end{array}\right]=\left[\begin{array}{c}7x_3\\ x_2\\ x_3\\ x_2\end{array}\right]$ by changing variables.

It means that when we transform the reduced echelon form to equations, we already have

$x_1=7x_3,x_4=x_2$

Or equivalently

$x_1-7x_3=0,x_2-x_4=0$

So the echelon form is easily seen to be

$\left[\begin{array}{cccc}1& 0& -7& 0\\ 0& 1& 0& -1\\ 0& 0& 0& 0\\ \end{array}\right]$

Since when we transform following matrix system to equations, we can attain above equations we want:

$\left[\begin{array}{cccc}1& 0& -7& 0\\ 0& 1& 0& -1\\ 0& 0& 0& 0\\ \end{array}\right]\cdot \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]$