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:

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

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;

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

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

Therefore, from the second equation, we have $z=2$.

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

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

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