Thread: Solving Linear Systems

1. Solving Linear Systems

use elementary row operations to reduce the given matrix to (a) row echelon form and (b) reduced row echelon form.

[-2 -4 7
-3 -6 10
1 2 -3]

2. Originally Posted by Yan
use elementary row operations to reduce the given matrix to (a) row echelon form and (b) reduced row echelon form.

[-2 -4 7
-3 -6 10
1 2 -3]
$\begin{bmatrix}-2&-4&7\\-3&-6&10\\1&2&-3\end{bmatrix}$

I will not show the matrices as I get it into REF, because I want you to put some effort into this; but I will give you the row operations that lead to REF. I leave it for you to discover the row operations to get it into RREF from REF.

1. $R_2\leftrightarrow R_3$

2. $3R_1+R_2\rightarrow R_2$

3. $2R_1+R_3\rightarrow R_3$

4. $-R_2+R_3\rightarrow R_3$

Can you take it from here?

--Chris

3. I got the answer

[1 2 -3
0 0 1
0 0 0]

Is that right?

4. Originally Posted by Yan
I got the answer

[1 2 -3
0 0 1
0 0 0]

Is that right?
Yes, that's the REF of the matrix.

Now what would the RREF look like?

--Chris

5. this is the part that i don't know how to do it. Can you help me to figure it out?

Thanks

6. Originally Posted by Yan
this is the part that i don't know how to do it. Can you help me to figure it out?

Thanks
It is actually one row operation.

Remember that in RREF, the leading 1's have zeros above and below them [except if the leading 1 is in the first row, first column--it would only have zeros below it].

I think this is enough of a hint for you to finish this off...

Can you try to take it from here?

--Chris

7. [1 2 0
0 0 1
0 0 0]
is it correct?

8. Originally Posted by Yan
[1 2 0
0 0 1
0 0 0]
is it correct?
Yes!

--Chris