Area of triangle using Determinants.

Question

I have a doubt with the following question.

Q: Using determinants, find the area of the triangle whose vertices are (1,2),(-2,3) and (-3,4).

Doubt: The question is very straightforward and needs the application of direct formula, but what I fail to understand is that I am again and again getting the same answer as "22" when keeping the COLUMN of 1's as the last column or as the first column. The formula says to keep it as the third one/last one but I am getting the same answer even when I kept it as the first column. Isn't it against the property of determinants of interchanging columns makes the sign to differ?. Kindly help.

Answer 1

Is your question really, why is $$\det\pmatrix{1&a&b\\1&c&d\\1&e&f} =\det\pmatrix{a&b&1\\c&d&1\\e&f&1}? $$ Swapping columns negates the determinant, so $$\det\pmatrix{1&a&b\\1&c&d\\1&e&f} =-\det\pmatrix{a&1&b\\c&1&d\\e&1&f} =\det\pmatrix{a&b&1\\c&d&1\\e&f&1}.$$

Answer 2

Suppose you have a square matrix $A$ with columns $c_1, c_2, \cdots, c_n$. Let me write the determinant as a function $d$, that gets the $n$ columns. With that notation, you are right that

$$d(c_1, \cdots, c_i, \cdots, c_j, \cdots, c_n) = -d(c_1, \cdots, c_j, \cdots, c_i, \cdots, c_n)$$

where you can see I only swapped columns $c_i$ and $c_j$. What you must note is that probably you swapped columns 2 times. You were calculating

$$d(o, c_1, c_2)$$

where $o$ is the column of ones, and you are probably comparing it to

$$d(c_1, c_2, o)$$

which has the same sign, given that you have to switch columns $2$ times:

$$d(o, c_1, c_2) = -d(c_2, c_1, o) = d(c_1, c_2, o)$$