# Having troubles putting matrices in another form, linear combination form i think.

1. Sep 16, 2005

### mr_coffee

Hello everyone
I started out with 2 matrices, which are 2 seperate problems but both want the same thing. It says Solve the system.
Here are the 2 problems:
Problems #14 and #15, i got answers for, but I don't know how to put it in that form.
for #14 i got:
-2x1 + x2 = 5;
0x1 + 0x2 = 0;

#15.

z = 7/22;
x+y + 5z = 1;
x+y = -13/22;

thanks.

2. Sep 16, 2005

### TD

For #14.

The equation you end up with is $- 2x_1 + x_2 = 5 \Leftrightarrow x_2 = 5 + 2x_1$

Let $x_2 = s$ then you have the set of solutions

$$V = \left\{ {\left[ {\begin{array}{*{20}c} {5 + 2s} \\ s \\ \end{array} } \right]|s \in \mathbb{R}} \right\}$$

You can write this as

$$\left[ {\begin{array}{*{20}c} {x_1 } \\ {x_2 } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 5 \\ 0 \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} 2 \\ 1 \\ \end{array} } \right]s$$

#15 is very similar, try that one

3. Sep 16, 2005

### mr_coffee

thank u so much, excellent explanation!

4. Sep 16, 2005

### mr_coffee

For number 15, I did the following:
$x_1 + x_2 +5x_2 = 1$
$x_3 = 7/22$
$x_1 + x_2 = -13/22$
$x_1 = -13/22 - x_2$
$x_2 = -13/22 - x_1$

$$V = \left\{ {\left[ {\begin{array}{*{20}c} {-13/22 - s} \\ {-13/22 - s} \\ { 0+ 7/22}\\ \end{array} } \right]|s \in \mathbb{R}} \right\}$$

$$\left[ {\begin{array}{*{20}c} {x_1 } \\ {x_2 } \\ {x_3} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} -13/22 \\ -13/22 \\ 0\\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} -1 \\ -1 \\ 7/22\\ \end{array} } \right]s$$

did i do that right or did i screw somthing up? Thanks.

5. Sep 17, 2005

### TD

Sorry, I was away for a while

It seems half-right... So we have:

$$\left\{ \begin{gathered} x + y + 5z = 1 \hfill \\ x + y = - 13/22 \hfill \\ z = 7/22 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} x + y = 1 - 5z = - 13/22 \hfill \\ x + y = - 13/22 \hfill \\ z = 7/22 \hfill \\ \end{gathered} \right$$

Now you see that the first 2 equations are the same, so we get to choose 1 (useful) variable, here either x or y. I'll choose y, so let $y = s$.

$$\left\{ \begin{gathered} x = - 13/22 - s \hfill \\ y = s \hfill \\ z = 7/22 \hfill \\ \end{gathered} \right$$

So we have the following solutions set

$$V = \left\{ {\left[ {\begin{array}{*{20}c} { - 13/22 - s} \\ s \\ {7/22} \\ \end{array} } \right]|s \in \mathbb{R}} \right\}$$

In parametric form this would give (watch where there is no s! z is indepedant of s!)

$$\left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} { - 13/22} \\ 0 \\ {7/22} \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} { - 1} \\ 1 \\ 0 \\ \end{array} } \right]s$$

6. Sep 18, 2005

### mr_coffee

Thanks alot!! sorry i'm really bad with these and the professor didn't explain jack

7. Sep 18, 2005

### TD

I hope it's more clear now, don't hesitate to ask for more details

8. Sep 18, 2005

### mr_coffee

Shiza...I just submitted that answer and it was wrong, So i went back and row reduced and I think I messed up. I can't row reduce any further then this can I?
|1 1 5 1|
|5 4 -2 -3|
-5R1 + R2 -> R2
|1 1 5 1 |
|0 -1 -27 -8|

Anything i do now just tkaes out a 0 and puts back a number. On the orignal problem I then thought I could do R1+R2.

So i'm left with
x+y+5z = 1
-y - 27z = -8
How do you choose what variable you let s to equal?
THanks.

9. Sep 18, 2005

### TD

Ah, I didn't check the original problems to see if your earlier work was correct.
So we have the initial problem:

$$\left\{ \begin{gathered} x + y + 5z = 1 \hfill \\ 5x + 4y - 2z = - 3 \hfill \\ \end{gathered} \right$$

In matrix-form:

$$\left( {\begin{array}{*{20}c} 1 & 1 & 5 & 1 \\ 5 & 4 & { - 2} & { - 3} \\ \end{array} } \right)$$

Now, what you did is correct but it's not finished yet, you can reduce more.
Normally, after full row reduction, you should get:

$$\left( {\begin{array}{*{20}c} 1 & 0 & { - 22} & { - 7} \\ 0 & 1 & {27} & 8 \\ \end{array} } \right)$$

Can you take it from here?

10. Sep 18, 2005

### mr_coffee

ahh thank you! this is what I did, look right?
image

11. Sep 19, 2005

### TD

Well, almost :)

Somewhere in the middle, you let $z = s$, so you get:

$$\left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} { - 7} \\ 8 \\ 0 \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} {22} \\ { - 27} \\ 1 \\ \end{array} } \right]s$$

12. Sep 19, 2005

### mr_coffee

ahh so close, how did you get a 1 for z?
In the equations, i never solved for z, i just let z = s, and z always had l ike 20 or 27 as a coefficent, thanks for the help@

13. Sep 19, 2005

### TD

Indeed, but because z = s, the coëfficiënt of s for z is 1, no? But there is no constant, hence the 0 is in the first column. It's not because you substitute z that it disappears...

14. Sep 19, 2005

### mr_coffee

OHhh!! Thank you for that explanation!!

15. Sep 19, 2005

No problem