# Gaussian Elimination

• March 31st 2009, 08:37 PM
skyslimit
Gaussian Elimination
Hey, can anyone help to explain a way to do Gaussian elimination by hand (not calc). I seem to get incorrect answers by hand..

For example, row operations for this matrix

[[ 4 -2 1 20]
[ 5 -4 -2 -28]
[- 4 -5 -4 6]]

Thanks :))
• April 1st 2009, 04:55 AM
Showcase_22
Quote:

$\begin{pmatrix}
{4}&{-2}&{1}&{20}\\
{5}&{-4}&{-2}&{-28}\\
{-4}&{-5}&{-4}&{5}
\end{pmatrix}$

The first step I would do is:

$\begin{pmatrix}
{1}&{-\frac{1}{2}}&{\frac{1}{4}}&{5}\\
{5}&{-4}&{-2}&{-28}\\
{-4}&{-5}&{-4}&{5}
\end{pmatrix}$

$R_1 \rightarrow \frac{1}{4}R_1$

$\begin{pmatrix}
{1}&{-\frac{1}{2}}&{\frac{1}{4}}&{5}\\
{0}&{-\frac{3}{2}}&{-\frac{13}{4}}&{-53}\\
{0}&{-7}&{-3}&{25}
\end{pmatrix}$

$R_1\rightarrow R_1$
$R_2 \rightarrow R_2-5R_1$
$R_3 \rightarrow R_3+4R_1$

$\begin{pmatrix}
{1}&{-\frac{1}{2}}&{\frac{1}{4}}&{5}\\
{0}&{1}&{\frac{13}{6}}&{\frac{106}{3}}\\
{0}&{-7}&{-3}&{25}
\end{pmatrix}$

$R_2 \rightarrow -\frac{2}{3}R_2$

$\begin{pmatrix}
{1}&{0}&{\frac{4}{3}}&{\frac{68}{3}}\\
{0}&{1}&{\frac{13}{6}}&{\frac{106}{3}}\\
{0}&{0}&{\frac{73}{6}}&{\frac{817}{3}}
\end{pmatrix}$

$R_1 \rightarrow R_1+\frac{1}{2}R_2$
$R_2 \rightarrow R_2$
$R_3 \rightarrow R_3+7R_2$

This is quite a complicated one! I'll leave the last steps for you to do. (Hint, it starts with $R_3 \rightarrow \frac{6}{73}R_3$.
• April 1st 2009, 05:21 AM
HallsofIvy
$\begin{bmatrix}4 & -2 & 1 & 20 \\ 5 & -4 & -2 & -28 \\ -4 & -5 & -4 & 6\end{bmatrix}$

We want to reduce the first three columns to the identity matrix and I recommend working one column at a time.

We want "1 0 0" in the first column. Since there is a "4" in first column, first row, multiply the first row by 1/4. Since there is a "5" in first column, second row, subtract 5 times the new first row from the second row. Since there is a "-4" in the first column, third row, add 4 times the new first row to the third row:

$\begin{bmatrix}1 & -\frac{1}{2} & \frac{1}{4} & 5 \\ 0 & -\frac{3}{2} & -\frac{13}{4} & -53 \\ 0 & -3 & -3 & 26\end{bmatrix}$

In order to get a "1" in second column, second row, we must multiply the second column by -2/3. Then add 1/2 the new second row to the first row and add 3 times the new second row to the third row:
$\begin{bmatrix}1 & 0 & \frac{4}{3} & \frac{68}{3} \\ 1 & 0 & \frac{13}{6} & \frac{106}{3} \\ 0 & 0 & \frac{4}{3} & 132\end{bmatrix}$

Finally, multiply the third row by 3/4 to get a "1" in the third column, third row, subtract 4/3 the new third row from the first row and subtract 13/6 the new third row from the second row:
$\begin{bmatrix}1 & 0 & 0 & \frac{226}{3} \\ 0 & 1 & 0 & -\frac{1075}{6} \\ 0 & 0 & 1 & 99\end{bmatrix}$