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 $x={\begin{bmatrix}7t\\s\\t\\s\end{bmatrix}}={\begin{bmatrix}7x_{3}\\x_{2}\\x_{3}\\x_{2}\end{bmatrix}}$ 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

$\color {blue}{\begin{bmatrix}1&0&-7&0\\0&1&0&-1\\0&0&0&0\\\end{bmatrix}}$

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

${\begin{bmatrix}1&0&-7&0\\0&1&0&-1\\0&0&0&0\\\end{bmatrix}}\cdot {\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}}={\begin{bmatrix}0\\0\\0\\0\end{bmatrix}}$