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4x4 Matrix Determinant watch

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    Hi, I am trying to do a 4x4 determinant using the quick approach I used previously for the determinant of 3x3 matrices.

    The quick approach is to sum the multiples of diagonal elements and minus the sum of the nondiagonal elements. This always gives me the corrent answer for 3x3 matrices.

    However, it doesn't do it for 4x4 matrices.

    For example (3x3):

    1 4 3
    5 2 1
    9 5 0

    Det (A) = (1*2*0+4*1*9+3*5*5) - (9*2*3+5*4*0+1*5*1) = 52 This is correct

    (4x4):

    1 -1 2 1
    2 1 -1 3
    1 5 -8 1
    4 5 -7 7

    Det (A) = (1*1*-8*7+-1*-1*1*4+2*3*1*5+1*2*5*-7)-(4*5*-1*1+1*1*2*7+2*-1*-7*1+1*5*-8*3) = 20 This is incorrect and the actual determinant is 0.

    What am I doing wrong, or what adjustment do I have to make for the 4x4 matrix? Does the quick method work for it?
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    (Original post by 41b)
    ...
    Determinants are defined recursively, eg. 3x3 determinants are defined in terms of 2x2 determinants, 4x4 determinants are defined in terms of 3x3 determinants.

    Your rule for 3x3 determinants works because the formula for a 2x2 determinant is nice and simple with only 2 terms. From each determinant, one term goes into the positive sum and one goes into the negative.
    It doesn't work for 3x3 matrices because the formula has 6 terms. There might be a similar rule but if so it's probably too hard to follow.

    The more general way of working out an n-dimensional determinant is by expanding along one row or column. For an nxn matrix A with elements a_i,j, \det(A) = \displaystyle\sum_{j=1}^n{(-1)^{i+j}a_{i,j}}\min(a_{i, j}), where \min(a_{i,j}) is the determinant of the matrix obtained by removing that row and column from A. Alternatively you can sum over one of the rows(summing over i).
    You can also perform row operations to make the calculations easier.

    For the example you've given(simplifying somewhat by row operations but leaving some for the sake of example),


    \begin{vmatrix}1&-1&2&1\\2&1&-1&3 \\ 1 & 5 & -8 & 1\\4&5&-7&7\end{vmatrix} =\begin{vmatrix}1&-1&2&1\\0&3&-5&1 \\ 0 & 6 & -10 & 0\\4&5&-7&7\end{vmatrix}
    =1(-1)^{1+1}\begin{vmatrix}3&-5&1\\6&-10&0\\5&-7&7\end{vmatrix} + 4(-1^{4+1})\begin{vmatrix}-1&2&1\\3&-5&1\\6&-10&0\end{vmatrix}

.

    You can then repeat the procedure for the 3x3s or use your rule to work this out.
    note: I could have reduced it to 1 3x3 determinant using row operations but I left 2 for the sake of the explanation.
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    (Original post by 41b)
    What am I doing wrong, or what adjustment do I have to make for the 4x4 matrix? Does the quick method work for it?
    A 4x4 determinant will involve the sum (including subtractions) of 24 terms, each of which is a product of 4 values. Personally, I wouldn't even attempt to find a formula in the same fashion as the 3x3 - clearly there is one, but I don't know whether it's easy to remember.

    In this case if you subtract the first, second and third rows from the last, it reduces the final row to 0,0,0,2. Then expand the determinant by that final row.
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    (Original post by 41b)
    Hi, I am trying to do a 4x4 determinant using the quick approach I used previously for the determinant of 3x3 matrices.

    The quick approach is to sum the multiples of diagonal elements and minus the sum of the nondiagonal elements. This always gives me the corrent answer for 3x3 matrices.

    However, it doesn't do it for 4x4 matrices.

    For example (3x3):

    1 4 3
    5 2 1
    9 5 0

    Det (A) = (1*2*0+4*1*9+3*5*5) - (9*2*3+5*4*0+1*5*1) = 52 This is correct

    (4x4):

    1 -1 2 1
    2 1 -1 3
    1 5 -8 1
    4 5 -7 7

    Det (A) = (1*1*-8*7+-1*-1*1*4+2*3*1*5+1*2*5*-7)-(4*5*-1*1+1*1*2*7+2*-1*-7*1+1*5*-8*3) = 20 This is incorrect and the actual determinant is 0.

    What am I doing wrong, or what adjustment do I have to make for the 4x4 matrix? Does the quick method work for it?
    Consider

    

\left(\begin{array}{cccc} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \\\end{array}\right)

    and its determinant

    

d g j m - c h j m - d f k m + b h k m + c f l m - b g l m - d g i n +  c h i n + d e k n - a h k n - c e l n + a g l n + d f i o - b h i o -  d e j o + a h j o + b e l o - a f l o - c f i p + b g i p +  c e j p - a g j p - b e k p + a f k p

    and see if you can spot the pattern. Note that the individual terms are selected one term from each row and column; then see if you can get the rule that assigns plus or minus. Then have a gin and tonic.
 
 
 
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