Tuesday, June 19, 2012

Calculate the determinant of this 6 x 6 matrix.{{1,1,0,0,0,0}, {2,2,2,0,0,0}, {0,3,3,3,0,0}, {0,0,4,4,4,0}, {0,0,0,5,5,5}, {0,0,0,0,6,6}}.

First we take out the factor 2 from the 2nd row, 3 from the 3rd row, 4 from the 4th row and 5 from the 5th row and 6 from the 6th row, the obtained factor is 6!(six factorial). The other factor is the determiminant of 6x6 order:{1,1,0,0,0,0},{1,1,1,0,0,0},{0,1,1,1,0,0},{0,0,1,1,1,0},{0,0,0,1,1,1} ,{0,0,0,0,1,1}.


Do an operation  of subtracting 1st row from 2nd row or R2 - R1 and the obtained determinant is:


{1,1,0,0,0,0},{0,1,1,0,0,0),{0,1,1,1,0,0},{0,0,1,1,1,0},{000111} ,{000011}.


Expand in terms of 1st column. The resultant 5x5 detrminant is:


{0,1,0,0,0},{1,1,1,0,0},{0,1,1,1,0},{0,0,1,1,1},{0,0,0,1,1}.


Expand in terms of 1st row, which has 2nd element1 and all other elements 0, as visible above.So the resulting determinant is (-1) multiplied by a detrminant of 4x4 order given below:


{1,1,0,0},{1,1,1,0},{0,1,1,1},{0,0,1,1}


We do the operation R2-R1 on this 4x4 detrminant and we get:


{1,1,0,0},{0,0,1,0},{0,1,1,1},{0,0,1,1}


Expand in terms  first column which is led by the element, 1 and followed by all  o elements. The resulting 3x3 detrminant is below:


{0,1,0},{1,1,1},{0,1,1}.


Expand by the 1st row, which has only 1 as 2nd element, the other two elements are 0's. So a (-1) factor and another factor , a  2x2 determinant, is as below:


{1,1},{1,1} the value of which is 1*1 - 1*1 = 0.


Recollect  all the factors and collect them together. The value of the original detminant is :


6!*(-1)*(-1)*0 and that is zero.


Six factorial times square of (minus one) times zero.

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