More fun in 2015, Part 29
M = ⎣ ⎢ ⎢ ⎡ 2 0 0 0 3 2 0 0 4 3 2 0 0 8 3 2 0 1 2 3 2 0 0 1 3 2 0 0 5 3 2 0 0 9 3 2 0 1 3 3 2 0 0 2 3 2 0 0 6 3 2 0 1 0 3 2 0 1 4 3 2 0 0 3 3 2 0 0 7 3 2 0 1 1 3 2 0 1 5 3 ⎦ ⎥ ⎥ ⎤
Find the determinant of the matrix above.
The answer is 5308416.
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1 solution
You can't have fun alone in 2015 :P . Click here . Haha
I think you meant to write a^3 in the first line where a is written in the ordered list ; as well as in the second line or am I mistaken? Also shouldn't the figure which you evaluated at the end , namely determinant of A , be negative?
We can write each entry of M as ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 = ( a , 3 a 2 , 3 a , 1 ) . ( 1 , b , b 2 , b 3 ) where a = 2 0 0 0 , 2 0 0 4 , 2 0 0 8 , 2 0 1 2 and b = 0 , 1 , 2 , 3 . Thus M = A B where the rows of A are ( a , 3 a 2 , 3 a , 1 ) and the columns of B are ( 1 , b , b 2 , b 3 ) . Note that B is a Vandermonde matrix with det ( B ) = ( 1 − 0 ) ( 2 − 0 ) ( 2 − 1 ) ( 3 − 0 ) ( 3 − 1 ) ( 3 − 2 ) = 1 2 . Now det ( A ) becomes a Vandermonde determinant after we factor out 3 ∗ 3 = 9 from the second and third column, with det ( A ) = 9 ( 2 0 0 4 − 2 0 0 0 ) ( 2 0 0 8 − 2 0 0 0 ) ( 2 0 0 8 − 2 0 0 4 ) ( 2 0 1 2 − 2 0 0 0 ) ( 2 0 1 2 − 2 0 0 4 ) ( 2 0 1 2 − 2 0 1 8 ) = 4 4 2 3 6 8 .
Finally, det ( M ) = ( det A ) ( det B ) = 1 2 ∗ 4 4 2 3 6 8 = 5 3 0 8 4 1 6