# Thread: linear algebra

1. ## linear algebra

[1 2 5 8
2 3 0 5
3 0 6 9
0 4 7 11]

ive tried 4 times to reduce this but my answers all look different and wrong. any suggestions?

2. Originally Posted by linearalgebra
[1 2 5 8
2 3 0 5
3 0 6 9
0 4 7 11]

ive tried 4 times to reduce this but my answers all look different and wrong. any suggestions?
Why not show some work. I got this

$\displaystyle \left[ \begin{array}{cccc} 1 & 0 & 0 &1 \\ 0 & 1 & 0 &1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 \end{array}\right]$

3. thanks..im done some row operations and right now my matrix looks like this
[1 -1 0 0
-1 7 1 7
1 0 2 3
0 0 1 1 ]

did you have that matrix at one point too, or should i just start over?
also, I have used about 10 row operations to get to that- is that too much? am i not seeing something really simple that could solve this using fewer row operations?
thanks.

4. okay did more operations and now im at the folowing...
[ 1 0 0 1
0 1 0 1
0 0 1 1
0 0 0 -2]

how can i change the last row to [0 0 0 0] without too many operations?

5. would it be valid if i multiplied the entire last row by 0?

6. Originally Posted by linearalgebra
thanks..im done some row operations and right now my matrix looks like this
[1 -1 0 0
-1 7 1 7
1 0 2 3
0 0 1 1 ]

did you have that matrix at one point too, or should i just start over?
also, I have used about 10 row operations to get to that- is that too much? am i not seeing something really simple that could solve this using fewer row operations?
thanks.
Your row reducing seems rather random. I think it's best to have an order. Starting with

$\displaystyle \left[ \begin{array}{cccc} 1 & 2 & 5 &8 \\ 2& 3 & 0 &5\\ 3 & 0 & 6 & 9\\ 0 & 4 & 7 & 11 \end{array}\right]$ target to get rid of the red numbers $\displaystyle \left[ \begin{array}{cccc} 1 & 2 & 5 &8 \\ \color{red}{2}& 3 & 0 &5\\ \color{red}{3} & 0 & 6 & 9\\ 0 & 4 & 7 & 11 \end{array}\right]$

$\displaystyle \begin{array}{c} 2R_1 -R_2 \rightarrow R_2\\ 3R_1- R_3 \rightarrow R_3 \end{array}$

$\displaystyle \left[ \begin{array}{cccc} 1 & 2 & 5 &8 \\ 0 & 1 & 10 &11\\ 0 & 6 & 9 & 15\\ 0 & 4 & 7 & 11 \end{array}\right]$ then use the second row to eliminate $\displaystyle \left[ \begin{array}{cccc} 1 & 2 & 5 &8 \\ 0 & 1 & 10 &11\\ 0 & \color{red}{6} & 9 & 15\\ 0 & \color{red}{4} & 7 & 11 \end{array}\right]$ etc

7. according to the way you did it, my next step looks liks this...
[ 1 2 5 8
0 1 10 11
0 0 51 51
0 0 33 33]

am i on the right track?

8. Originally Posted by linearalgebra
according to the way you did it, my next step looks liks this...
[ 1 2 5 8
0 1 10 11
0 0 51 51
0 0 33 33]

am i on the right track?
Yes, excellent! Now dividing the last two rows by 51 an 33 respectively, we can then start working your way up.