How do you find the determinant of ((14, -13, 0), (3, 8, -1), (-10, -2, 5))?

1 Answer
Mar 4, 2016

det=597

Explanation:

There are two ways of evaluating a third-order determinant, but the most easiest is by;

|(a_11quada_12quada_13), (a_21quada_22quada_23), (a_31quada_32quada_33)|=a_11|(a_22quada_23), (a_32quada_33)|-a_12|(a_21quada_23), (a_31quada_33)|+a_13|(a_21quada_22), (a_31quada_32)|

det=a_11(a_22a_33-a_23a_32)-a_12(a_21a_33-a_23a_31)+a_13(a_21a_32-a_22a_31)

What needs to be remembered is that we can choose any row to break down the third-order arrangement into second-order arrangement, and it is easier when choosing a row with 0 in it because any second-order arrangement multiplied by 0 is 0. but also we must consider its sign;

|(a_11quada_12quada_13), (a_21quada_22quada_23), (a_31quada_32quada_33)|=|(+ - +), (- + -), (+ - +)|

When breaking down for example into a_11|(a_22quada_23), (a_32quada_33)|, the |(a_22quada_23), (a_32quada_33)| is determined by eliminating the number that horizontally and vertically in line with a_11. By this way, a_12,a_13,a_21 and a_31 is eliminated.

So, from the question given, we break down it by using row 1 since it has 0 in it;

|(14, -13, 0), (3, 8, -1), (-10, -2, 5)|=

=14|(8, -1), (-2, 5)|-(-13)|(3, -1), (-10, 5)|+0|(3, 8), (-10, -2)|

det=14[8(5)-(-1)(-2)]+13[3(5)-(-1)(-10)]+0[3(-2)-8(-10)]

det=14(38)+13(5)+0(74)

det=532+65

det=597