Question

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

Answer 1

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`