How do you find the determinant of #((3, 5, 0, 6), (2, 3, 2, 0), (2, 4, 0, 7), (-3, 2, 2, 3))#?

1 Answer
Jun 10, 2016

Answer:

I found #80#

Explanation:

I would try to simplify a bit our original matrix to get a line or column of all zeroes but one number.

One column that seems good in this sense is the third one where we have two zeroes already!

Now, I can multiply the second line by #-1# and add it to the fourth line to get the other zero!

So;
the second line would become:
#-2,-3-2,0#
we add this line to the third and we get:
#((3,5,0,6),(2,3,color(red)(2),0),(2,4,0,7),(-5,-1,0,3))#

we see that the third column has 3 zeroes and only one number: we can cancel the column and line crossing at #2# (red) to reduce the order of our matrix BUT we have to keep the #2# we used to do this with the appropriate sign!

Remember that:
#((+,-,+,-),(-,+,color(red)(-),+),(+,-,+,-),(-,+,-,+))#

so our #2# must have the corresponding minus sign and we get:
#color(red)(-2)((3,5,6),(2,4,7),(-5,-1,3))# as reduced matrix!

The determinant can be now "easily" evaluated using Laplace or Sarrus to get:
#"Det"=-2[3(12+7)-5(6+35)+6(-2+20)]=-2[57-205+108]=80#