Solving Systems of Equations: #14 & #15

[mr_coffee]

Hello everyone
I started out with 2 matrices, which are 2 separate problems but both want the same thing. It says Solve the system.
Here are the 2 problems:
http://img141.imageshack.us/img141/8382/matrix9ek.jpg #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.

 

[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

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

You can write this as

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

#15 is very similar, try that one

 

[mr_coffee]

thank u so much, excellent explanation!

 

[mr_coffee]

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


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


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

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

 

[TD]

Sorry, I was away for a while

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

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

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 $y = s$.

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

So we have the following solutions set

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

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

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

 

[mr_coffee]

Thanks a lot! sorry I'm really bad with these and the professor didn't explain jack

 

[TD]

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

 

[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.

 

[TD]

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

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

In matrix-form:

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

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

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

Can you take it from here?

 

[mr_coffee]

ahh thank you! this is what I did, look right?
http://show.imagehosting.us/show/687576/0/nouser_687/T0_-1_687576.jpg

 

[TD]

Well, almost :)

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

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

 

[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 coefficient, thanks for the help@

 

[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...

 

[mr_coffee]

OHhh! Thank you for that explanation!

 

[TD]

No problem