Results 1 to 4 of 4

Thread: Determinant of a 3X3 matrix

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    19

    Determinant of a 3X3 matrix

    I am really confused and frustrated i put this matrix in a calculator and found the determinant to be 9 but when i do the math myself i get -11? What am i doing wrong here?? Here is the matrix:

    1 -2 1
    2 2 -1
    1 -1 2

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    5
    bet you show your work and we'll spot what you're doing wrong.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    15
    det\left(\begin{array}{ccc}1&-2&1\\2&2&-1\\1&-1&2\end{array}\right)
     = (1*2*2) + (-2*-1*1) + (1*2*-1) - (1*2*1) - (-1*-1*1) - (2*2*-2) = 9 by using the diagonals method.
    to do this, write like this:
    det\left|\begin{array}{ccc}1&-2&1\\2&2&-1\\1&-1&2\end{array}\right|\begin{array}{cc}1&-2\\2&2\\1&-1\end{array}

    You multiply the first element by the one on the down sloping diagonal, and so forth.
    element (1,1) * element (2,2) * element (3,3)
    element (2,1) * element (3,2) * element (4,3)
    and so on.

    When you go up a diagonal, you subtract the product from the total.

    Hope this helps!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    13,707
    Thanks
    537
    That method, unfortunately, only works for 3 by 3 determinants.

    Another way to do it- expand by minors on the first row:
    \left|\begin{array}{ccc}1 & - 2 & 1 \\ 2 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right|= 1 \left|\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right|-  (-2)\left|\begin{array}{cc}2 & -1 \\ 1 & 2\end{array}\right| + 1\left|\begin{array}{cc}2 & 2 \\ 1 & -1\end{array}\right|

    Yet another- row reduce:
    \begin{pmatrix}1 & -2 & 1 \\ 2 & 2 & -1\\ 1 & -1 & 2\end{pmatrix}

    Subtract twice the first row from the second row and subtract the first row from the third row to get
    \begin{pmatrix}1 & -2 & 1 \\ 0 & 6 & -3 \\ 0 & 1 & 1\end{pmatrix}

    Swap the second and third rows, then subtract 6 times that new second row from the new third row to get
    \begin{pmatrix}1 & -2 & 1 \\ 0 & 1 & 1 \\0 & 0 &-9\end{pmatrix}

    Adding (or subtracting) a multiple of one row from another does not change the determinant and swapping two rows multiplies the determinant by -1 so the determinant of the original matrix is -1 times the determinant of this "upper triangular matrix".

    (Multiplying or dividing one row by a number, a row operation not used here, multiplies of divides the determinant by that number.)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Derivative of a matrix inverse and matrix determinant
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 24th 2011, 08:18 AM
  2. Determinant of a matrix
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: November 10th 2010, 09:07 PM
  3. determinant of this 4x4 matrix
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 2nd 2010, 07:02 PM
  4. Determinant of matrix
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 25th 2010, 06:35 PM
  5. Determinant of a 2 x 3 matrix?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 6th 2008, 01:19 PM

Search Tags


/mathhelpforum @mathhelpforum