How do you find the determinant of #((1, 2, 1), (0, 4, 3), (1, 2, 2))#?

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Answer 1 · 2016-03-11T17:08:53

#det=4#

#D=1|(4,3),(2,2)| -2|(0,3),(1,2)|+1|(0,4),(1,2)|#

#=1(8-6)-2(0-3)+1(0-4)#-use Cramer's rule #|(r,s),(t,u)| = ru-st#

#=2+6-4#
#det=4#

Answer 2 · 2016-03-11T23:55:09

Use alternative method to find determinant is #4#

Use two properties of determinants:

The determinant of a matrix is unchanged by adding any multiple of a row to another row.
The determinant of an upper triangular matrix is the product of the diagonal.

Given:

#((1, 2, 1), (0, 4, 3), (1, 2, 2))#

Subtract the first row from the third (i.e. add #-1# times the first row to the third) to get:

#((1, 2, 1), (0, 4, 3), (0, 0, 1))#

The determinant of this upper triangular matrix is the product of the diagonal:

#1 xx 4 xx 1 = 4#