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

  • Thread starter mr_coffee
  • Start date
  • #1
1,629
1
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:
http://img141.imageshack.us/img141/8382/matrix9ek.jpg [Broken] #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.
 
Last edited by a moderator:

Answers and Replies

  • #2
TD
Homework Helper
1,022
0
For #14.

The equation you end up with is [itex] - 2x_1 + x_2 = 5 \Leftrightarrow x_2 = 5 + 2x_1 [/itex]

Let [itex]x_2 = s[/itex] then you have the set of solutions

[tex]V = \left\{ {\left[ {\begin{array}{*{20}c}
{5 + 2s} \\
s \\

\end{array} } \right]|s \in \mathbb{R}} \right\}
[/tex]

You can write this as

[tex]\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[/tex]

#15 is very similar, try that one :smile:
 
  • #3
1,629
1
thank u so much, excellent explanation!
 
  • #4
1,629
1
For number 15, I did the following:
[itex] x_1 + x_2 +5x_2 = 1 [/itex]
[itex]x_3 = 7/22 [/itex]
[itex]x_1 + x_2 = -13/22[/itex]
[itex]x_1 = -13/22 - x_2[/itex]
[itex]x_2 = -13/22 - x_1[/itex]


[tex]V = \left\{ {\left[ {\begin{array}{*{20}c}
{-13/22 - s} \\
{-13/22 - s} \\
{ 0+ 7/22}\\

\end{array} } \right]|s \in \mathbb{R}} \right\}
[/tex]


[tex]\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[/tex]

did i do that right or did i screw somthing up? Thanks.
 
  • #5
TD
Homework Helper
1,022
0
Sorry, I was away for a while :smile:

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

[tex]\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[/tex]

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 [itex]y = s[/itex].

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

So we have the following solutions set

[tex]V = \left\{ {\left[ {\begin{array}{*{20}c}
{ - 13/22 - s} \\
s \\
{7/22} \\

\end{array} } \right]|s \in \mathbb{R}} \right\}[/tex]

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

[tex]\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[/tex]
 
  • #6
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1
Thanks alot!! sorry i'm really bad with these and the professor didn't explain jack
 
  • #7
TD
Homework Helper
1,022
0
I hope it's more clear now, don't hesitate to ask for more details :smile:
 
  • #8
1,629
1
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
TD
Homework Helper
1,022
0
Ah, I didn't check the original problems to see if your earlier work was correct.
So we have the initial problem:

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

In matrix-form:

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

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

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

Can you take it from here?
 
  • #11
TD
Homework Helper
1,022
0
Well, almost :)

Somewhere in the middle, you let [itex]z = s[/itex], so you get:

[tex]\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[/tex]
 
  • #12
1,629
1
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
TD
Homework Helper
1,022
0
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
1,629
1
OHhh!! :biggrin: Thank you for that explanation!!
 
  • #15
TD
Homework Helper
1,022
0
No problem :smile:
 

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