1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
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 - Determinants

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data

    If [tex]det\left[
    \begin {array}{ccc}
    a&1&d\\
    \noalign{\medskip}
    b&1&e\\
    \noalign{\medskip}
    c&1&f
    \end {array}
    \right]=-4
    [/tex] and [tex]det\left[
    \begin {array}{ccc}
    a&1&d\\
    \noalign{\medskip}
    b&2&e\\
    \noalign{\medskip}
    c&3&f
    \end {array}
    \right]=-1
    [/tex],

    then [tex]det\left[
    \begin {array}{ccc}
    a&8&d\\
    \noalign{\medskip}
    b&8&e\\
    \noalign{\medskip}
    c&8&f
    \end {array}
    \right]=___
    [/tex]

    and [tex]det\left[
    \begin {array}{ccc}
    a&-1&d\\
    \noalign{\medskip}
    b&-4&e\\
    \noalign{\medskip}
    c&-7&f
    \end {array}
    \right]=___
    [/tex]

    3. The attempt at a solution

    For the first question, I'm pretty sure that I can factor out an 8 as it is a scalar multiple of the second column. One of the properties of determinants is that if a row or column is multiplied by a scalar, then we can factor the scalar out and then multiply the determinant by that scalar. Thus, the answer would be -32.
    However, I am really stumped by the second question. I am sure that the scalar -1 is somehow multiplied into the matrix, but I am not sure how the numbers were obtained. I'm thinking that each row must have resulted from the scalar multiple of another row. However, if that's the case, then why are the unknowns unaffected?
    I am seriously at my wit's end and any direction would be helpful.
    Thank you!
     
  2. jcsd
  3. Oct 10, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    (-1,-4,-7) is (-3)*(1,2,3)+2*(1,1,1). A determinant is a linear function of each of it's columns.
     
  4. Oct 10, 2008 #3

    Defennder

    User Avatar
    Homework Helper

    Note that detA = detA^T. And also the determinant of A+B where A,B differ by a single row is det(A+B). Now multiply the column 1,2,3 of that matrix by 3. What do you have to multiply the 2nd column of the 1,1,1 column matrix such that adding up both gives you -1,-4,-7?
     
  5. Oct 10, 2008 #4
    Wow! It makes so much more sense now that you've mentioned detA=detA^T to me! Thank you so much!
     
  6. Oct 10, 2008 #5
    Now that you have already seen the algebraic explanation, here is an image tutorial showing the geometric intuition behind it:
    http://img137.imageshack.us/img137/6679/determinantrowopsjx2.png

    The image explains in 2D, but in higher dimensions everything is the same except parallelograms become paralellipipeds and the determinant measures volume not area.

    Can you see why the determinant of a singular matrix has to be zero?
     
  7. Oct 11, 2008 #6

    Defennder

    User Avatar
    Homework Helper

    Uh, what has that got to do with the OP's question?
     
  8. Oct 11, 2008 #7
    It is a visual proof that
    1) Multiplying a column by a constant scales the determinant by that constant, and
    2) The determinant is linear in any particular column.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear Algebra - Determinants
  1. Linear algebra (Replies: 3)

  2. Linear Algebra (Replies: 5)

  3. Linear Algebra (Replies: 1)

Loading...