I am reading this text and I'm stuck on how they get to the reduced row echelon form of this matrix:
I've tried a ton of combinations but I still can't get there. Can someone show me the way? Here's one that I have but then I'm stuck:
$\begin{matrix}
\\ 1 & 2 & 5
\\ 0 & -5 & -5
\end{matrix}$
But then how do I get rid of the 2 in the second column of the first row to 0?
Best Answer
We have $$\begin{bmatrix} 1 & 2 &5 \\ 2 & -1 & 5 \end{bmatrix}$$ Denoting row $i$ as $r_i$ and assume that the left arrow $A \leftarrow B$ means that we are putting quantity $B$ in $A$. Moreover, we know that we can do linear operations on the rows to reach the row-echelon form, i.e.
$r_2 \leftarrow r_2 - 2r_1$
$$\begin{bmatrix} 1 & 2 &5 \\ 0 & -5 & -5 \end{bmatrix}$$
$r_1 \leftarrow r_1 + \frac{2}{5}r_2$
$$\begin{bmatrix} 1 & 0 &3 \\ 0 & -5 & -5 \end{bmatrix}$$ $r_2 \leftarrow -\frac{1}{5}r_2$ $$\begin{bmatrix} 1 & 0 &3 \\ 0 & 1 & 1 \end{bmatrix}$$