Determinant evaluating!
M = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 7 1 3 1 9 2 5 3 1 2 8 1 4 2 0 2 6 3 2 3 9 1 5 2 1 2 7 3 3 4 1 0 1 6 2 2 2 8 3 4 5 1 1 1 7 2 3 2 9 3 5 6 1 2 1 8 2 4 3 0 3 6 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
Evaluate the matrix determinant above.
The answer is 0.
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2 solutions
Moderator note:
Bingo! Would this be true for all n × n matrix if the matrix is of the form below?
M = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 n + 1 2 n + 1 ⋅ ⋅ ⋅ 2 n + 2 ⋅ ⋅ ⋅ ⋅ 3 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ n 2 − 1 n 2 n ⋅ ⋅ ( n − 1 ) n n 2 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
Yes or will always be true
Since the first element ( a 1 1 ) is 1 , we can use Chió's rule, which states a new matrix, with the same determinant as the first and order one unit lesser, can be achieved through the following graphical transformation:
M = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 7 1 3 1 9 2 5 3 1 2 8 1 4 2 0 2 6 3 2 3 9 1 5 2 1 2 7 3 3 4 1 0 1 6 2 2 2 8 3 4 5 1 1 1 7 2 3 2 9 3 5 6 1 2 1 8 2 4 3 0 3 6 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
M = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 8 − 2 . 7 1 4 − 2 . 1 3 2 0 − 2 . 1 9 2 6 − 2 . 2 5 3 2 − 2 . 3 1 9 − 3 . 7 1 5 − 3 . 1 3 2 1 − 3 . 1 9 2 7 − 3 . 2 5 3 3 − 3 . 3 1 1 0 − 4 . 7 1 6 − 4 . 1 3 2 2 − 4 . 1 9 2 8 − 4 . 2 5 3 4 − 4 . 3 1 1 1 − 5 . 7 1 7 − 5 . 1 3 2 3 − 5 . 1 9 2 9 − 5 . 2 5 3 5 − 5 . 3 1 1 2 − 6 . 7 1 8 − 6 . 1 3 2 4 − 6 . 1 9 3 0 − 6 . 2 5 3 6 − 6 . 3 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
M = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ − 6 − 1 2 − 1 8 − 2 4 − 3 0 − 1 2 − 2 4 − 3 6 − 4 8 − 6 0 − 1 8 − 3 6 − 5 4 − 7 2 − 9 0 − 2 4 − 4 8 − 7 2 − 9 6 − 1 2 0 − 3 0 − 6 0 − 9 0 − 1 2 0 − 1 5 0 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
And from determinants properties, like d e t ( c . M ) = c n . d e t ( M ) , where c is a scalar, n is the order of the matrix and M is the matrix in question, we can make it becomes:
( − 6 ) 6 . M = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 2 3 4 5 2 4 6 8 1 0 3 6 9 1 2 1 5 4 8 1 2 1 6 2 0 5 1 0 1 5 2 0 2 5 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
Applying Chió's rule once again, we get that the determinant is the same that the given by the fourth order matrix below:
( − 6 ) 6 . M = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
Since the determinant of a null matrix is zero, hence, the answer is M = 0 .
We can easily see that if we subtract elements of row 1 by row 2 and elements of row 2 by row 3 we get a new determinant with two rows equal. So by properties of determinants the determinant equates to zero