Results 1 to 10 of 10

Math Help - Linear Equation and matrices.

  1. #1
    Junior Member
    Joined
    Oct 2011
    Posts
    36

    Linear Equation and matrices.

    It's been a ten years ago when I learning those equations, some how recently my boss just give me a question and asking for help, I d't have any ideal back to carry out those equation where all the knowledge already give back to teacher, i guess.

    Therefore, hope some one here can give me a little help on it. Thank you.

    Part 1
    Writing those in linear equation,
    a) x-y+z = 3
    b) 4x-3y-z = 6
    c) 3x +y+2z = 23

    in the form of Ax=b. State the size of matrix A and its main diagonal elements

    Thanks to Prove It Part 1 Solved

    Part 2
    Using the elementary row operation (ERO) on the augmented matrix A|b, solve the system of linear equation above.



    Linear Equation and matrices.-1.jpg <---Try to resolve but not correct, need help
    Last edited by gilagila; October 2nd 2011 at 02:45 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293

    Re: Linear Equation

    Quote Originally Posted by gilagila View Post
    It's been a ten years ago when I learning those equations, some how recently my boss just give me a question and asking for help, I d't have any ideal back to carry out those equation where all the knowledge already give back to teacher, i guess.

    Therefore, hope some one here can give me a little help on it. Thank you.

    Writing those in linear equation,
    a) x-y+z = 3
    b) 4x-3y-z = 6
    c) 3x +y+2z = 23

    in the form of Ax=b. State the size of matrix A and its main diagonal elements
    \displaystyle \mathbf{A} is the matrix which consists of all the coefficients of your variables, \displaystyle \mathbf{x} is a column vector containing your variables, and \displaystyle \mathbf{b} is a column vector containing the values on the right-hand-side of the equation.

    So do you think you can write down what values are in the matrices?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2011
    Posts
    36

    Re: Linear Equation

    Quote Originally Posted by Prove It View Post
    \displaystyle \mathbf{A} is the matrix which consists of all the coefficients of your variables, \displaystyle \mathbf{x} is a column vector containing your variables, and \displaystyle \mathbf{b} is a column vector containing the values on the right-hand-side of the equation.

    So do you think you can write down what values are in the matrices?
    Not sure is it correct or not, let me try some
    a) x-y+z = 3
    b) 4x-3y-z = 6
    c) 3x +y+2z = 23

    if A = 1 -1 1...........x = x.........b= 3
    .........4 -3 -1 ..............y..............6
    .........3 1 2...................z............23

    so Size of matrix A = square matrix 3 x 3 and the main diagonal elements = 1, -3 , 2

    am i right ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293

    Re: Linear Equation

    Quote Originally Posted by gilagila View Post
    Not sure is it correct or not, let me try some
    a) x-y+z = 3
    b) 4x-3y-z = 6
    c) 3x +y+2z = 23

    if A = 1 -1 1 x = x b= 3
    4 -3 -1 y 6
    3 1 2 z 23

    so Size of matrix A = square matrix 3 x 3 and the main diagonal elements = 1, -3 , 2

    am i right ?
    I know what you have done, just to make it more readable...

    \displaystyle \mathbf{A} = \left[\begin{matrix}1 & -1 & \phantom{-}1 \\ 4 & -3 & -1 \\ 3 & \phantom{-}1 & \phantom{-}2\end{matrix}\right]

    \displaystyle \mathbf{x} = \left[\begin{matrix} x \\ y \\ z \end{matrix}\right]

    \displaystyle \mathbf{b} = \left[\begin{matrix} 3 \\ 6 \\ 23 \end{matrix}\right]

    And you are correct that the elements in the main diagonal are 1, -3 and 2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2011
    Posts
    36

    Re: Linear Equation

    Quote Originally Posted by Prove It View Post
    I know what you have done, just to make it more readable...

    \displaystyle \mathbf{A} = \left[\begin{matrix}1 & -1 & \phantom{-}1 \\ 4 & -3 & -1 \\ 3 & \phantom{-}1 & \phantom{-}2\end{matrix}\right]

    \displaystyle \mathbf{x} = \left[\begin{matrix} x \\ y \\ z \end{matrix}\right]

    \displaystyle \mathbf{b} = \left[\begin{matrix} 3 \\ 6 \\ 23 \end{matrix}\right]

    And you are correct that the elements in the main diagonal are 1, -3 and 2
    new in this forum, just want to ask how you type those matrix ? using what function / icon in the edit area ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,312
    Thanks
    692

    Re: Linear Equation

    it's called latex, there is a tutorial here. to use it, you use the "tex" tags. for example:

    "[tex ]x^2 + x + \sqrt{2}[ /tex]" produces x^2 + x + \sqrt{2}

    (don't put the spaces after the tex or before the /tex)

    when row-reducing, there are 3 operations we can perform:

    1) switch two rows,
    2) multiply one row by a non-zero number,
    3) multiply one row by a non-zero number and add/subtract it from another row.

    the idea is try to try get a leading 1 in each row, each leading 1 is rightward of the rows above, and we want all 0's below each leading 1 (you can also try to have 0's above each leading 1 as well, which brings the matrix into reduced row echelon form). we may wind up with 0 rows at the bottom.

    in your matrix, since we already have a leading 1 in the first row, the first step is to eliminate the rest of the first column. subtracting 4 times the 1st row from the second, and then subtracting 3 times the first row form the third, we get:

    \begin{bmatrix}1&-1&1&3\\0&1&-5&-6\\0&4&-1&14 \end{bmatrix}

    (this is the augmented matrix) can you continue?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2011
    Posts
    36

    Re: Linear Equation

    Quote Originally Posted by Deveno View Post
    it's called latex, there is a tutorial here. to use it, you use the "tex" tags. for example:

    "[tex ]x^2 + x + \sqrt{2}[ /tex]" produces x^2 + x + \sqrt{2}

    (don't put the spaces after the tex or before the /tex)

    when row-reducing, there are 3 operations we can perform:

    1) switch two rows,
    2) multiply one row by a non-zero number,
    3) multiply one row by a non-zero number and add/subtract it from another row.

    the idea is try to try get a leading 1 in each row, each leading 1 is rightward of the rows above, and we want all 0's below each leading 1 (you can also try to have 0's above each leading 1 as well, which brings the matrix into reduced row echelon form). we may wind up with 0 rows at the bottom.

    in your matrix, since we already have a leading 1 in the first row, the first step is to eliminate the rest of the first column. subtracting 4 times the 1st row from the second, and then subtracting 3 times the first row form the third, we get:

    \begin{bmatrix}1&-1&1&3\\0&1&-5&-6\\0&4&-1&14 \end{bmatrix}

    (this is the augmented matrix) can you continue?
    thanks for the help. I just do it in MS Word and make it as a picture, somehow d't know wher eis wrong, cannot get the answer right. Pls advice. TQ
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2011
    Posts
    36

    Re: Linear Equation

    Quote Originally Posted by Deveno View Post

    in your matrix, since we already have a leading 1 in the first row, the first step is to eliminate the rest of the first column. subtracting 4 times the 1st row from the second, and then subtracting 3 times the first row form the third, we get:

    \begin{bmatrix}1&-1&1&3\\0&1&-5&-6\\0&4&-1&14 \end{bmatrix}

    (this is the augmented matrix) can you continue?
    when 4R1 - R2, in r2 it should be 4(1)-4, 4(-1)-(-3), 4(1)-(-1), 4(3)-6 = 0,-1,5,6, why your no.2 figure is +1 not -1 ? and your no.4 figure is -6 not +6 ?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,312
    Thanks
    692

    Re: Linear Equation

    there are different ways to row-reduce, you used slightly different operations than i did, but you can see that you (eventually) wound up with the same 2nd row as me, and the negative of the 3rd row.

    the error you made in your row-reduction is in the very last step, instead of adding 8 to -3 (in the 4th column), you added 8 to 3, getting 11, instead of the 5 you should have gotten.

    you may check that x = 5, y = 4, z = 2 is indeed a solution to your original equations.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Oct 2011
    Posts
    36

    Re: Linear Equation

    Quote Originally Posted by Deveno View Post
    there are different ways to row-reduce, you used slightly different operations than i did, but you can see that you (eventually) wound up with the same 2nd row as me, and the negative of the 3rd row.

    the error you made in your row-reduction is in the very last step, instead of adding 8 to -3 (in the 4th column), you added 8 to 3, getting 11, instead of the 5 you should have gotten.

    you may check that x = 5, y = 4, z = 2 is indeed a solution to your original equations.
    oops..ya..finally i got the answer. Thanks for the tip on the mistake on last row deduction.

    This problem has been solved. Thank you all.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. matrices of linear equations
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: December 30th 2010, 05:14 AM
  2. Linear combinations - matrices
    Posted in the Advanced Algebra Forum
    Replies: 11
    Last Post: September 18th 2010, 05:25 PM
  3. Matrices and linear transformations
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 22nd 2010, 04:21 AM
  4. Replies: 2
    Last Post: April 25th 2009, 07:40 AM
  5. Linear Equation Using Matrices
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 1st 2007, 09:09 PM

/mathhelpforum @mathhelpforum