Question

How do you row reduce this? No matter what I try I get ugly numbers.

\[ \left[\begin{matrix} 22 & -16 & -2 \\ -16 & 12 & 0 \end{matrix}\right] \]

The answer is
\[ \left[\begin{matrix} 1 & 0 & -3 \\ 0 & 1 & -4 \end{matrix}\right] \]

Answer 1

\[\begin{bmatrix}22&-16&-2\\-16&12&0\end{bmatrix}\]
I'll use \(R_i\) to denote the operations I'd use on row \(i\), and \(R_i\to R_j\) to denote which row I'll replace after each operation.

First thing I would do would be to divide everything by the gcf, in effect doing \(\dfrac{1}{2}R_1\to R_1\) and \(\dfrac{1}{4}R_2\to R_2\):
\[\begin{bmatrix}11&-8&-1\\-4&3&0\end{bmatrix}\]
Following, I'd have \(\dfrac{11}{4}R_2+R_1\to R_2\):
\[\begin{bmatrix}11&-8&-1\\-11&\frac{33}{4}&0\end{bmatrix}\]
\(R_1+R_2\to R_2\):
\[\begin{bmatrix}11&-8&-1\\0&\frac{1}{4}&-1\end{bmatrix}\]
\(4R_2\to R_2\):
\[\begin{bmatrix}11&-8&-1\\0&1&-4\end{bmatrix}\]
\(8R_2+R_1\to R_1\):
\[\begin{bmatrix}11&0&-33\\0&1&-4\end{bmatrix}\]
And finally, \(\dfrac{1}{11}R_1\to R_1\):
\[\begin{bmatrix}1&0&-3\\0&1&-4\end{bmatrix}\]

Answer 2

@SithsAndGiggles Oh wow thank you! I had a very similar result about half way through, i might have messed up a negative sign somewhere though.