Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear Algebra -finding determinant of a matrix

  1. Oct 7, 2014 #1
    the first row 1 0 0 2
    the 2nd row 0 1 2 0
    the 3rd row 0 2 1 0
    the 4th row 2 0 0 1

    I would like to ask which is the most efficient way of solving this ques.Though i can solve but is long method, I know there must have some quick 1, appreciate if u can share it. thank you
     
  2. jcsd
  3. Oct 7, 2014 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Expansion using minors is your best approach.
     
  4. Oct 8, 2014 #3
    minors?
     
  5. Oct 8, 2014 #4

    Mark44

    Staff: Mentor

    If aij is an element in matrix A, the minor of this element, Aij, is the submatrix that contains all of the elements of A that are not in row i or in column j.
    For example, a11 in your matrix is 1. The minor of this element is A11, which is
    \begin{bmatrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}

    For more details of how this works to calculate the determinant, do a web search for "determinant minor".
     
  6. Dec 30, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Expanding by minors on the first row:
    [tex]\left|\begin{array}{cccc}1 & 0 & 0 & 2 \\ 0 & 1 & 2 & 0 \\ 0 & 2 & 1 & 0 \\ 2 & 0 & 0 & 1 \end{array}\right|= 1\left|\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|- 2 \left|\begin{array}{ccc}0 & 0 & 2 \\ 1 & 2 & 0 \\ 2 & 1 & 0 \end{array}\right|[/tex].
    You could then expand the first of those two determinants on the last row and the second on the first row.

    You can also use "row reduction" to reduce the underlying matrix to a triangular matrix. The determinant of a triangular matrix is just the product of the numbers on the diagonal and "row reduction" changes the determinant is a regular way. The basic "row operation" are
    1) multiply every number on a row by the same number. This multiplies the determinant by that number.
    2) swap two rows. This multiplies the determinant by -1.
    3) add a multiple of a row to another row. This does not change the determinant- and it is always possible to reduce a matrix to a triangular matrix using only this.

    Here, if you add -2 times the first row to the last row you get [tex]\begin{bmatrix}1& 0 & 0 & 2 \\ 0 & 1 & 2 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & -3\end{bmatrix}[/tex].

    Now, add -2 times the second row to the third row: [tex]\begin{bmatrix}1 & 0 & 0 & 2 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & -3 & 0 \\ 0 & 0 & 0 & -3\end{bmatrix}[/tex]

    Since we used only "add a multiple of one row to another" that matrix has the same determinant as the original.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linear Algebra -finding determinant of a matrix
Loading...