Question

For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Answer 1

Answer:

The augmented matrix of the system is $\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]$.

We apply operations rows:

1. We swap row 1 and 2 and obtain the matrix $\left[\begin{array}{cccc}1&2&1&4\\2&3&-1&14\\5&9&2&7\end{array}\right]$.

2. Of the above matrix we subtract row 1 from row 2 twice (R2 - 2R1) and we subtract row 1 from row 3, 5 times. (R3-5R1). We obtain the matrix $\left[\begin{array}{cccc}1&2&1&4\\0&-1&-3&6\\0&-1&-3&-13\end{array}\right]$

3. Of the above matrix we subtract row 2 from row1 twice (R3 - R2) and multiply the row 1 by -1 (-R2). Weobtain the matrix $\left[\begin{array}{cccc}1&2&1&4\\0&1&3&-6\\0&0&0&-19\end{array}\right]$.

Since each pivote is an one then we conclude that the above matrix is the reduced row-echelon form of the matrix of the system.