Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 08/Solution 1

The given system of the linear equations can be represented by the augmented matrix:

${\displaystyle \left[{\begin{array}{ccc|c}1&1&1&5\\2&2&0&6\\0&2&4&8\end{array}}\right]}$

Since elementary row operations do not change the solution to the linear system, we remove the entry in column 1 of row 2 by subtracting off -2 times the first row;

${\displaystyle \left[{\begin{array}{ccc|c}1&1&1&5\\2&2&0&6\\0&2&4&8\end{array}}\right]{\stackrel {\longrightarrow }{R_{2}-2R_{1}}}\left[{\begin{array}{ccc|c}1&1&1&5\\0&0&-2&-4\\0&2&4&8\end{array}}\right]}$

which is equivalent to ${\displaystyle {\begin{cases}x+y+z=5\\-2z=-4\\2y+4z=8\end{cases}}}$

Therefore, from the second equation, we have ${\displaystyle z=2}$.

And the third equation implies ${\displaystyle 2y=8-4z=8-8=0}$, so ${\displaystyle y=0}$.

Finally, from the first equation, we obtain ${\displaystyle x=5-y-z=3}$.

Therefore, the solution is ${\displaystyle \color {blue}x=3,y=0,z=2}$.