Question

How to find determinant of 6 by 6 matrix?

Answer 1

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.