Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By HallsofIvy

Math Help - Help Linear Dependent - Shortcut

  1. #1
    Junior Member
    Joined
    Sep 2012
    From
    NY
    Posts
    32

    Help Linear Dependent - Shortcut

    Show that {x2 , x2+x+1, x2+x, x-1} is LD

    I took like 20 mins to find the solution with trial and error to get

    (x2+x) - 1/2(x2+x+1) -1/2(x-1) -1/2(x2) =0 ---> LD

    Are there any formula or shortcut for this?


    How about {x4+x2+1, x4-x2+1, x4-x2-1}?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644

    Re: Help Linear Dependent - Shortcut

    Hello, angelme!

    \text{Show that: }\:\begin{Bmatrix}x^2 \\ x^2+x+1 \\ x^2+x \\ x-1\end{Bmatrix}\,\text{ are linearly dependent.}

    We want nonzero numbers a,b,c,d so that:

    . . a(x^2) + b(x^2+x+1) + c(x^2+x) + d(x - 1) \;=\;0

    . . ax^2 + bx^2 + bx + b + cx^2 + cx + dx - d \;=\;0

    . . (a+b+c)x^2 + (b+c+d)x + (b-d) \;=\;0x^2 + 0x + 0


    Two polynomials are equal if their corresponding coefficients are equal:
    . . . . \begin{Bmatrix} a+b+c \;=\;0 \\ b+c+d \;=\;0 \\ b-d \;=\;0 \end{Bmatrix}

    Solve the system: . \begin{Bmatrix}a&=& t \\ b &=& t \\ c &=&\text{-}2t \\ d &=& t \end{Bmatrix}
    Last edited by Soroban; October 7th 2012 at 09:59 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2012
    From
    NY
    Posts
    32

    Re: Help Linear Dependent - Shortcut

    Thank you Soraban!!!

    So...for {x
    4+x2+1, x4-x2+1, x4-x2-1}

    a(
    x4+x2+1) + b(x4-x2+1) + c(x4-x2-1) =0
    (a+b+c)
    x4 + (a-b-c)x2 + (a+b-c) = 0

    then I get

    a=0
    b=0
    c=0

    Does it mean it is LI?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2012
    From
    NY
    Posts
    32

    Re: Help Linear Dependent - Shortcut

    someone please help correct my work above

    Question 2: {x4+x2+1, x4-x2+1, x4-x2-1}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,696
    Thanks
    1467

    Re: Help Linear Dependent - Shortcut

    Quote Originally Posted by angelme View Post
    Thank you Soraban!!!

    So...for {x
    4+x2+1, x4-x2+1, x4-x2-1}

    a(
    x4+x2+1) + b(x4-x2+1) + c(x4-x2-1) =0
    (a+b+c)
    x4 + (a-b-c)x2 + (a+b-c) = 0

    then I get

    a=0
    b=0
    c=0

    Does it mean it is LI?
    Yes.

    Yet another way to do this is to simply choose three numbers for x, getting three equations for a, b, and c.

    For example, taking x= 0, the a(x^4+ x^2+ 1)+ b(x^4+ x^2- 1)+ c(x^4- x^2- 1)= 0 becomes a- b- c= 0. Taking x= 1, 3a+ b- c= 0. Taking x= 2, 21a+ 19b+ 11c= 0.
    Last edited by HallsofIvy; October 9th 2012 at 08:05 AM.
    Thanks from angelme
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Independent and Dependent
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 24th 2010, 04:27 AM
  2. Linear dependent/independent
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 7th 2009, 06:51 PM
  3. Linear dependent vectors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 24th 2009, 04:36 PM
  4. Another linear dependent/independent question
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: March 27th 2009, 06:00 PM
  5. Vectors Linear Dependent?
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 23rd 2008, 03:20 AM

Search Tags


/mathhelpforum @mathhelpforum