1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Sep 16, 2005 #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:
    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. jcsd
  3. Sep 16, 2005 #2

    TD

    User Avatar
    Homework Helper

    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:
     
  4. Sep 16, 2005 #3
    thank u so much, excellent explanation!
     
  5. Sep 16, 2005 #4
    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.
     
  6. Sep 17, 2005 #5

    TD

    User Avatar
    Homework Helper

    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]
     
  7. Sep 18, 2005 #6
    Thanks alot!! sorry i'm really bad with these and the professor didn't explain jack
     
  8. Sep 18, 2005 #7

    TD

    User Avatar
    Homework Helper

    I hope it's more clear now, don't hesitate to ask for more details :smile:
     
  9. Sep 18, 2005 #8
    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.
     
  10. Sep 18, 2005 #9

    TD

    User Avatar
    Homework Helper

    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. Sep 18, 2005 #10
    ahh thank you! this is what I did, look right?
    image
     
  12. Sep 19, 2005 #11

    TD

    User Avatar
    Homework Helper

    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]
     
  13. Sep 19, 2005 #12
    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@
     
  14. Sep 19, 2005 #13

    TD

    User Avatar
    Homework Helper

    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...
     
  15. Sep 19, 2005 #14
    OHhh!! :biggrin: Thank you for that explanation!!
     
  16. Sep 19, 2005 #15

    TD

    User Avatar
    Homework Helper

    No problem :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Having troubles putting matrices in another form, linear combination form i think.
Loading...