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How do I simplify 4x4 det before solving? I am doing somethingwrong

  1. Nov 23, 2012 #1
    1. The problem statement, all variables and given/known data

    solve the det
    3 1 1 0
    -4 5 0 -1
    0 7 2 8
    2 -1 0 3

    2. Relevant equations



    3. The attempt at a solution
    I am trying to solve it and get a different answer each time so I am doing something I shouldn't obviously.

    Ok I am adding rows to other rows
    (ex 2*row1 +row2)
    Same I do for columns.

    When I expend the matrix I pay attention to the + and -
    So what is wrong :S

    This is what I did (I didn't choose the shortest way because I wanted to make sure I am solving it right but hence something is off :S)

    I did 2*R4+r2,R3-2*R1
    I opened the det by a3,1
    And got to this by this point:
    1*
    -4 5 -1​
    -6 5 8​
    0 3 5 ​
    then I did
    R1-R2,3*R1+R2
    then opened it by a2,1
    I got the det:
    6*
    5 -19​
    3 5​
    Which is 492 (should b 82)
    Now in all the cases I tried I reached 82 before multiplying the number outside the 2x2 det (6 in this case)


    Well it is kinda hard showing the steps of the det though it is a simple one but mainly is there something wrong with the steps I take?
     
  2. jcsd
  3. Nov 23, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Okay, that's a good start. It gets a 0 in the first column, second row.

    But I don't understand why you did that? I thought you were trying to get 0s in the first column. This gives you
    [tex]\begin{bmatrix}3 & 1 & 1 & 0 \\ 0 & 3 & 0 & 5 \\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3\end{bmatrix}[/tex]

    Perhaps I misunderstood what you meant before. Because you already have two 0s in the third column, if you subtract 2 times the first row from the third row, Replacing R3 with R3- 2R1, you get
    [tex]\left|\begin{array}{cccc}3 & 1 & 1 & 0 \\-4 & 5 & 0 & -1\\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3 \end{array}\right|[/tex]
    Expanding on the third column, you get
    [tex]1\left|\begin{array}{ccc}4 & 5 & -1 \\ 6 & 5 & 8 \\ 2 & -1 & 3\end{array}\right|[/tex]
    Although I still don't see how you got that third row.

     
  4. Nov 23, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Okay, that's a good start. It gets a 0 in the first column, second row.

    But I don't understand why you did that? I thought you were trying to get 0s in the first column. This gives you
    [tex]\begin{bmatrix}3 & 1 & 1 & 0 \\ 0 & 3 & 0 & 5 \\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3\end{bmatrix}[/tex]

    Perhaps I misunderstood what you meant before. Because you already have two 0s in the third column, if you subtract 2 times the first row from the third row, Replacing R3 with R3- 2R1, you get
    [tex]\left|\begin{array}{cccc}3 & 1 & 1 & 0 \\-4 & 5 & 0 & -1\\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3 \end{array}\right|[/tex]
    Expanding on the third column, you get
    [tex]1\left|\begin{array}{ccc}4 & 5 & -1 \\ 6 & 5 & 8 \\ 2 & -1 & 3\end{array}\right|[/tex]
    Although I still don't see how you got that third row.

     
  5. Nov 23, 2012 #4
    Not everything I did had to have a meaning I was playing around with it.
    I figured out what I was doing wrong though...I treated it as matrix .
    When I write 2R2+R3
    I actually mean R2=2R2+R3
    I didn't know I can't do that with dets or I change its value:S
     
  6. Nov 24, 2012 #5
    Row operations of course do change the value of the determinant.
    Why don't you try expansion by minors to solve it? Might be a bit tedious, but would work after all.
     
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