Question

Solve the system of equations by transforming a matrix representing the system of equation into reduced row echelon form.

Answer 1

Take the augmented matrix,

$\left[\begin{array}{ccc|c}2&1&-3&-20\\1&2&1&-3\\1&-1&5&19\end{array}\right]$

Swap the row 1 and row 2:

$\left[\begin{array}{ccc|c}1&2&1&-3\\2&1&-3&-20\\1&-1&5&19\end{array}\right]$

Add -2(row 1) to row 2, and -1(row 1) to row 3:

$\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&-3&4&22\end{array}\right]$

Add -1(row 2) to row 3:

$\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&9&36\end{array}\right]$

Multiply through row 3 by 1/9:

$\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&1&4\end{array}\right]$

Add 5(row 3) to row 2:

$\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&0&6\\0&0&1&4\end{array}\right]$

Multiply through row 2 by -1/3:

$\left[\begin{array}{ccc|c}1&2&1&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]$

Add -2(row 2) and -1(row 3) to row 1:

$\left[\begin{array}{ccc|c}1&0&0&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]$

So we have $\boxed{x=-3,y=-2,z=4}$.