# Row reduction question, is this right?

$$\begin{bmatrix}-1 & 2 & -5\2 & -1 & 6\ 2 & -2 & 7\end{bmatrix}$$

What i did was multiply the first line by 2 = $$\begin{bmatrix}-1 & -2 & -5\end{bmatrix} * 2 = \begin{bmatrix}-2 & 4 & -10\end{bmatrix}$$ So here i added the first line multiplied by 2 to the second and third ones to create zeroes $$\begin{bmatrix}-1 & -2 & -5\end{bmatrix}$$ $$\begin{bmatrix}2 & -1 & 6\end{bmatrix} + \begin{bmatrix}-2 & 4 & -10\end{bmatrix}$$ $$\begin{bmatrix}2 & -2 & 7\end{bmatrix}+ \begin{bmatrix}-2 & 4 & -10\end{bmatrix}$$

Resulting in $$\begin{bmatrix}-1 & 2 & -5\0 & 3 & 4\ 0 & 2 & 3\end{bmatrix}$$

Finally i changed this second line so i could eliminate the third line's y variable:

$$\begin{bmatrix}0 & 3 & 4\end{bmatrix} * 2/3 = \begin{bmatrix}0 & 6/3 & 8/3\end{bmatrix}$$

As 6/3 equals 2

$$\begin{bmatrix}0 & 3 & 4\end{bmatrix} * 2/3 = \begin{bmatrix}0 & 2 & 8/3\end{bmatrix}$$

and with this line i could finish it:

$$\begin{bmatrix}0 & 2 & 9/3\end{bmatrix} - \begin{bmatrix}0 & 2 & 8/3\end{bmatrix} = \begin{bmatrix}0 & 0 & 1/3\end{bmatrix}$$

So i'll end up with

$$\begin{bmatrix}-1 & 2 & -5\0 & 3 & 4\ 0 & 0 & 1/3\end{bmatrix}$$

Monday, 25 November 2013 18:39 GMT