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Math Help - Matrix again

  1. #1
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    Matrix again

    Hi I really need help!

    I did the questions but something went wrong in my working, can someone please show me how to derive the answer with row reduction?! I really am confused when to express my answer in parametric equations!



    \begin{bmatrix} 1 & 4 & 11 \\1 & 6 & 17 \\1 & 4 & 8 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix} =


    = \begin{bmatrix} 7 \\9 \\4 \end{bmatrix}

    After Row reduction I got



    \begin{bmatrix} 1 & 4 & 11 \\0 & 2 & 6 \\0 & 0 & -3 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix}[/tex] =


    \begin{bmatrix} 7 \\2 \\-3 \end{bmatrix}


    For Row two, I took row 2 - 2 times row 1
    FOr Row 3, I took row 3 minus Row 1

    So, since I did not get all zeros ir at least three zeros in the bottom entries, I resorted to represent unique solutions in parametric equations, I got y=1-3t, z=t and x=3+t..however, the answer is in whole numbers for each x, y and z. And when I pressed the matrix into the calculator to solve for x, y and z, I got whole numbers too..

    How do I know that I am to use parametric equations to solve? And how do I get the whole numbers in this case?

    Thank you so much, your help is much appreciated!
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  2. #2
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    Re: Matrix again

    first of all, you didn't "finish" row-reduction, because you only reduced your matrix to upper-triangular form (and not row echelon, or reduced row echelon form, which makes the solutions more transparent).

    but look, your last row means that:

    0x + 0y - 3z = -3 that is:

    -3z = -3
    z = 1 <--not a parameter, a VALUE.

    your second row means:

    0x + 2y + 6z = 2 that is:

    2y + 6z = 2

    but we already know that z = 1! so this becomes:

    2y + 6 = 2
    2y = -4
    y = -2 <---again, not a parameter, an actual number.

    finally, your top row means:

    x + 4y + 11z = 7

    since y = -2, ans z = 1, this is:

    x - 8 + 11 = 7
    x + 3 = 7
    x = 4

    *************************

    the "other" question you ask, how do you know "when to use parameters?" depends on the RANK of your original matrix.

    row-reduction is one way we find out the rank, after row-reduction, it is the number of non-zero rows remaining. in this case, the rank of the original matrix is 3.

    this means several things, all of which are equivalent:

    1) the matrix A in your original problem is invertible
    2) the matrix A is non-singular (the null space is ONLY the 0-vector (0,0,0)) that is A(x,y,z) = (0,0,0) implies x = y = z = 0.
    3) the column space of A has dimension 3 (it is thus ALL of R3).
    4) the columns of A are linearly independent
    5) the equation A(x,y,z) = (7,9,4) has a UNIQUE solution
    6) the determinant of A is non-zero
    7) we can use Cramer's Rule to solve the system of equations
    Thanks from topsquark
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  3. #3
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    Re: Matrix again

    Hello, Tutu!

    There's an error in your first step . . .


    \begin{bmatrix} 1 & 4 & 11 \\1 & 6 & 17 \\1 & 4 & 8 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix} \;=\; \begin{bmatrix} 7 \\9 \\4 \end{bmatrix}

    \text{We have: }\;\left[\begin{array}{ccc|c} 1&4&11&7 \\ 1&6&17 & 9 \\ 1&4&8&4 \end{array}\right]

    \begin{array}{c}\\R_2-R_1 \\ R_3-R_1 \end{array}\;\left[\begin{array}{ccc|c}1&4&11 & 7 \\ 0&2&6&2 \\ 0&0&\text{-}3 & \text{-}3 \end{array}\right]

    . . \begin{array}{c}\\ \;\frac{1}{2}R_2 \\ \text{-}\frac{1}{3}R_3 \end{array}\;\;\left[\begin{array}{ccc|c}1&4&11&7 \\ 0&1&3&1 \\ 0&0&1&1 \end{array}\right]

    \begin{array}{c}R_1-4R_2 \\ R_2-3R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c}1&0&\text{-}1 & 3 \\ 0&1&0&\text{-}2 \\ 0&0&1&1 \end{array}\right]

    \begin{array}{c}R_1+R_3 \\ \\ \\ \end{array}\;\; \left[\begin{array}{ccc|c}1&0&0&4 \\ 0&1&0&\text{-}2 \\ 0&0&1&1 \end{array}\right]

    \text{Therefore: }\:\begin{Bmatrix}x &=& 4 \\ y &=& \text{-}2 \\ z &=& 1 \end{Bmatrix}
    Thanks from topsquark
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  4. #4
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    Re: Matrix again

    i don't see any "error", the upper triangular matrix you obtain is the same one he did. this is sufficient to solve the system of equations, since a linear equation in one unknown over a field always has a solution, unless it is of the form:

    0x + b = c, with b and c unequal. since the last row is non-zero, that is not the case here.
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  5. #5
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    Re: Matrix again

    Thank you so much! I understand it!

    I tried it on another question, I got the only 1 out of 3 answers, could you help me to derive the next few?

    \begin{bmatrix} 1 & 1 & 1 \\2 & 4 & 1 \\2 & 3 & 1 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix} = \begin{bmatrix} 6 \\5 \\6 \end{bmatrix}

    I reduced it to
    \begin{bmatrix} 1 & 1 & 1 \\0 & 1 & 0 \\0 & -1 & 0 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix} = \begin{bmatrix} 6 \\-1 \\1 \end{bmatrix}

    So I got y=-1.
    However, as I tried to further reduce the matrix like Soroban did, I cannot seem to get the values for x and z individually, I could only get x+z = 7.

    Please help me out in solving for x and z!

    Thank you so so so much both!
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  6. #6
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    Re: Matrix again

    again, you don't seem to understand what it is you're doing when you are row-reducing. your matrix is NOT:

    1)upper triangular

    2)upper row-echelon form

    3)reduced row-echelon form


    it appears in this case you have also row-reduced incorrectly. i can obtain the same last row you did by taking R3-R2, but its not kosher to then subtract R3 from R2.

    instead, you need to subtract 2*R1 from R2, giving:

    \begin{bmatrix}1&1&1\\0&2&-1\\0&-1&0 \end{bmatrix} \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}6\\-7\\1 \end{bmatrix}

    now we can add (1/2)*R2 to R3, and we get:

    \begin{bmatrix}1&1&1\\0&2&-1\\0&0&-\frac{1}{2} \end{bmatrix} \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}6\\-7\\ -\frac{5}{2} \end{bmatrix}

    again, we have a matrix of rank 3, so we don't need parameters.

    the last equation is:

    -z/2 = -5/2

    so z = 5

    the second equation is now:

    2y - z = -7
    2y - 5 = -7
    2y = -2
    y = -1

    the first equation is now:

    x + y + z = 6
    x - 1 + 5 = 6
    x = 2

    we can check this:

    2 - 1 + 5 = 6 (check!)
    2(2) + 4(-1) + 5 = 4 - 4 + 5 = 5 (check!!)
    2(2) + 3(-1) + 5 = 4 - 3 + 5 = 6 (check!!!)
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  7. #7
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    Re: Matrix again

    Thank you again!

    May I clarify, prior to the first question I asked, how is \begin{bmatrix} 1 & 4 & 11 \\0 & 2 & 6 \\0 & 0 & -3 \end{bmatrix} not a row echelon?

    From what I know, row echelon is when there should be zeros below leading ones, which exists here..
    Doe sthat mean that each time I do row reduction, I either change it to role echelon OR reduced row echelon?

    Thank you!
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  8. #8
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    Re: Matrix again

    it's not in row echelon form because you don't have leading 1's...for example the first non-zero entry in row 2 is 2, not 1, and the first non-zero entry in row 3 is -3, not 1.

    to put it in row echelon form, you would have to multiply row 2 by 1/2, and row 3 by -1/3.

    to put it in REDUCED row echelon form (rref), you would need to clear out the entries ABOVE the leading 1's as well.
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