# Difference between revisions of "12-240/Classnotes for Tuesday November 15"

$A=\begin{pmatrix}0&2&4&2&2\\4&4&4&8&0\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}$.
Solution. Using (invertible!) row/column operations we aim to bring $A$ to look as close as possible to an identity matrix:
 Do Get 1. Bring a $1$ to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by $1/4$. $\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}$ 2. Add $(-8)$ times the first row to the third row, in order to cancel the $8$ in position 3-1. $\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\6&3&2&9&1\end{pmatrix}$ 3. Likewise add $(-6)$ times the first row to the fourth row, in order to cancel the $6$ in position 4-1. $\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}$ 4. With similar column operations (you need three of those) cancel all the entries in the first row (except, of course, the first, which is used in the canceling). $\begin{pmatrix}1&0&0&0&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}$ 5. Turn the 2-2 entry to a $1$ by multiplying the second row by $1/2$. $\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}$ 6. Using two row operations "clean" the second column; that is, cancel all entries in it other than the "pivot" $1$ at position 2-2. $\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}$ 7. Using three column operations clean the second row except the pivot. $\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}$ 8. Clean up the row and the column of the $4$ in position 3-3 by first multiplying the third row by $1/4$ and then performing the appropriate row and column transformations. Notice that by pure luck, the $4$ at position 4-5 of the matrix gets killed in action. $\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\end{pmatrix}$