1. ## Linear Algebra

Hello,

I am doing my math homework, and I come across two problems I can't tend to do. Something always come up wrong. If anyone can help me, it is truly appreciated. Thank you so much.

I have to find the reduce echelon form for these two problems

$\displaystyle A=$\left( \begin{array}{ccccc}1 & -2 & 1 & 1 & 4\\2 & 1 & -3 & -1 & 6\\1 & -7 & -6 & 2 & 6\end{array} \right)$$

$\displaystyle A=$\left( \begin{array}{ccccc}0 & 1 & 2 & 1\\2 & 1 & 0 & 2\\0 & 2 & 1 & 1\end{array} \right)$$

2. Originally Posted by jenjen
Hello,

I am doing my math homework, and I come across two problems I can't tend to do. Something always come up wrong. If anyone can help me, it is truly appreciated. Thank you so much.

I have to find the reduce echelon form for these two problems

$\displaystyle A=$\left( \begin{array}{ccccc}1 & -2 & 1 & 1 & 4\\2 & 1 & -3 & -1 & 6\\1 & -7 & -6 & 2 & 6\end{array} \right)$$
$\displaystyle A=$\left( \begin{array}{ccccc} 1 & -2 & 1 & 1 & 4\\ 2 & 1 & -3 & -1 & 6\\ 1 & -7 & -6 & 2 & 6 \end{array} \right)$$

Subtract twice the first row from the second, and subtract the first row from
the third:

$\displaystyle B=$\left( \begin{array}{ccccc} 1 & -2 & 1 & 1 & 4\\ 2-2 & 1+4 & -3-2 & -1-2 & 6-8\\ 1-1 & -7+2 & -6-1 & 2-1 & 6-4 \end{array} \right)$$$\displaystyle =$\left( \begin{array}{ccccc} 1 & -2 & 1 & 1 & 4\\ 0 & 5 & -5 & -3 & -2\\ 0 & -5 & -7 & 1 & 2 \end{array} \right)$$

Now add the second row to the last row:

$\displaystyle C=$\left( \begin{array}{ccccc} 1 & -2 & 1 & 1 & 4\\ 0 & 5 & -5 & -3 & -2\\ 0 & -5+5 & -7-5 & 1-3 & 2-2 \end{array} \right)$$]$\displaystyle =$\left( \begin{array}{ccccc} 1 & -2 & 1 & 1 & 4\\ 0 & 5 & -5 & -3 & -2\\ 0 & 0 & -12 & -2 & 0 \end{array} \right)$$

Now finally divide the second row through by $\displaystyle 5$, and the third row by $\displaystyle -12$ to get:

$\displaystyle $\left( \begin{array}{ccccc} 1 & -2 & 1 & 1 & 4\\ 0 & 1 & -1 & -3/5 & -2/5\\ 0 & 0 & 1 & 1/6 & 0 \end{array} \right)$$

RonL