1. Let's evaluate the determinant:
Remember DET|a1 b1 c1|
|a2 b2 c2|
|a3 b3 c3| = a1(b2c3 - c2b3) - b1(a2c3 - c2a3) + c1(a2b3 - b2a3)
Using this we can calucluate our determinant as:
(x+1)(x-2)(x+3) -0 + 0 = (x+1)(x-2)(x+3) so we straight away see that the only value of x listed for which this is NOT zero is 1. Therefore the answer is a. The link provided by George confirms this.
2. Lets just expand the determinant again using the same formula.
(x-1)[(x-2)(x+3) + 6] + 2(-x - 3 + 6) - 1(-3 -3x + 6) =
x^3 + x^2 - 6x + 6x - x^2 - x + 6 - 6 - 2x - 6 + 12 + 3 + 3x - 6 =
x^3 + 3
So the answer is a or d
Last edited by gnitsuk (2006-03-14 20:37:27)