How do you find the determinant of #((5, 3, 1, 2), (0, 1, -1, 3), (2, 7, -4, 1), (3, 3, 5, -2))#?

1 Answer
May 21, 2016

Answer:

See below.

Explanation:

Because the question is a 'how' question I'm going to explain how to compute this determinant but leave the actual computation up to the reader.

The best way to do it, is to start in the top left hand corner, pick the number occupying this position, ignore the numbers in its row and column, then multiplying that top left number by the determinant of the numbers remaining (I'll show this below). You then go along the top row, alternating between positive and negative (always starting with positive), ignoring the row and column we are on and then multiplying your selected number by the determinant of the remaining numbers.

Here's the demonstration:
(For the purpose of this answer I will use #||# symbols to refer to determinants)

#|((5,3,1,2),(0,1,-1,3),(2,7,-4,1),(3,3,5,-2))|#

Becomes:

#5|((1,-1,3),(7,-4,1),(3,5,-2))|-3|((0,-1,3),(2,-4,1),(3,5,-2))|+1|((0,1,3),(2,7,1),(3,3,-2))|-2|((0,1,-1),(2,7,-4),(3,3,5))|#

Taking 3x3 determinants is the same thing. For the purpose of this I'll just demonstrate taking the first one:

#|((1,-1,3),(7,-4,1),(3,5,-2))| = 1|((-4,1),(5,-2))|-(-1)|((7,1),(3,-2))|+3|((7,-4),(3,5))|#

I'll finally just show the general rule for taking a 2x2 determinant:

#|((a,b),(c,d))|=ad-bc#