How do you find the determinant of #((3, -3, 3), (1, -2, -1), (2, 0, 6))#?
1 Answer
Nov 7, 2016
Explanation:
Expanding using row1 we have;
# = (3){(-2)(6)-(0)(-1)} + (3){(1)(6)-(2)(-1)} + (3){(1)(0)-(2)(-2)} #
# = (3)(-12) + (3)(6+2) + (3)(0-(-4)) #
# = -36 + (3)(8) + 12 #
# = -36 + 24 + 12 #
# = 0 #