Results 1 to 3 of 3

Math Help - Help on a Linear Algebra Problem

  1. #1
    Newbie
    Joined
    Jul 2009
    Posts
    8

    Help on a Linear Algebra Problem

    Hi All,

    I got this problem and it seems like it should be straight forward, but I get the wrong answer. I listed below my reasoning. What is wrong with what I am doing? Any help would be greatly appreciated.

    Let B={ 2x^3-5x^2+3x+7,x^2-4x+9,1/2x+5,3}

    Find < 3x^3+6x^2-8x+27> by method of comparing coefficients.
    Here is what I did.
    < 3x^3+6x^2-8x+2> =C1< 2x^3-5x^2+3x+7> +C2< x^2-4x+9> +C3< 1/2x+5>+C4*3

    2C1=3, C1=3/2
    C2-5=6 C2= 11
    1/2C3-4+3= -8
    1/2C3=-7
    C3=-14
    3C4+9+7+5=27
    3C4=6
    C4=2
    <3/2,11, -14,2>
    Last edited by egshih; July 19th 2009 at 11:36 AM. Reason: Testing out LaTex
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by egshih View Post
    Hi All,

    I got this problem and it seems like it should be straight forward, but I get the wrong answer. I listed below my reasoning. What is wrong with what I am doing? Any help would be greatly appreciated.

    Let B={ 2x^3-5x^2+3x+7,x^2-4x+9,1/2x+5,3}

    Find < 3x^3+6x^2-8x+27> by method of comparing coefficients.
    Here is what I did.
    < 3x^3+6x^2-8x+2> =C1< 2x^3-5x^2+3x+7> +C2< x^2-4x+9> +C3< 1/2x+5>+C4*3

    2C1=3, C1=3/2
    C2-5=6 C2= 11
    1/2C3-4+3= -8
    1/2C3=-7
    C3=-14
    3C4+9+7+5=27
    3C4=6
    C4=2
    <3/2,11, -14,2>
    You should be using the value of C1 in your computation of C2 as b_1 has a second degree term, ...

    You could try it this way:

    p(x)=3x^3+6x^2-8x+27

    Then as the leading coefficient of b_1 is 2 we have:

    p(x)=\frac{3}{2}b_1(x)+\frac{27}{2}x^2-8x+12

    similarly as the leading coefficient of b_2 is 1 we have:

    p(x)=\frac{3}{2}b_1(x)+\frac{27}{2}b_2(x)+46x-\frac{219}{2}

    etc.

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    Quote Originally Posted by egshih View Post
    Hi All,

    I got this problem and it seems like it should be straight forward, but I get the wrong answer. I listed below my reasoning. What is wrong with what I am doing? Any help would be greatly appreciated.

    Let B={ 2x^3-5x^2+3x+7,x^2-4x+9,1/2x+5,3}

    Find < 3x^3+6x^2-8x+27> by method of comparing coefficients.
    Here is what I did.
    < 3x^3+6x^2-8x+2> =C1< 2x^3-5x^2+3x+7> +C2< x^2-4x+9> +C3< 1/2x+5>+C4*3
    Better if you go ahead and do the combination on the right:
    3x^3+6x^2-8x+2= C1x^3+(-5C1+C2)x^2+ (3C1- 4C2 + (1/2)C3)x+ (7C1+9C2+ 5C3+ 3C4)

    2C1=3, C1=3/2
    Yes, that is correct.

    C2-5=6 C2= 11
    -5C1+ C2= -5(3/2)+ C2= C2- 15/2= 11. C2= 11+ 15/2= 37/2.
    You need -5C1, not just -5.

    1/2C3-4+3= -8
    1/2C3=-7
    C3=-14
    3C4+9+7+5=27
    3C4=6
    C4=2
    <3/2,11, -14,2>
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Algebra Problem
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 4th 2009, 02:36 PM
  2. Linear algebra problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 6th 2009, 11:49 PM
  3. linear algebra problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 8th 2009, 05:23 PM
  4. Last Linear Algebra problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 9th 2008, 10:06 AM
  5. Linear Algebra problem
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: February 13th 2007, 05:13 PM

Search Tags


/mathhelpforum @mathhelpforum