You can prove, by hand, that the determinany of a 2x2 (and a 3x3 matrix) represents the effect on volumes the linear map has.
In general it is a little more tricky to prove, and but can be explained thusly:
given a vector space V there is another vector space \Lambda^2V (which I shall for typesetting reasons label /\^2(V)) called the alternating or antisymmetric square. If e_i is a basis of V then /\^2(V) then has as basis the symbols e_i/\e_j subject to the relation that e_i/\e_j=-e_j/\e_i. That is swapping over the arguments changes sign, hence the name alternating. There are other rules too such as the symbol (e_i+e_j)/\e_k = e_i/\e_k + e_j/\e_k.
Now, I can repeat this construction and define /\^n(V) by taking as a basis the 'n-fold wedge', so this has basis the symbols where we have n terms not two, but with the same rules that if we swap over two arguments we change the sign.
What happens if n=dimV? I claim that the resutling space /\^n(V) is one dimensional. To see this notice that if any wedge has a repeated element then it is identified with the zero vector (in /\^2 where it is easier to write e/\f=-f/\e, so if I let e=f, then e/\e=-e/\e, so e/\e=0). What this means is that because of the rule about swapping things round, given e_i/\e_j/\../\e_k I can swap them round so that the subscripts are all increasing, and because if I repeat a subscript I get zero then for non-zero terms these subscripts must be strictly increasing. If n=dimV how can choose a strictly increasing set of length n from the numbers 1,..,n? In exactly one way, so there is exactly one basis vector.
With n=dimV, what /\^n represents is volume. If you take n vectors, write them in terms of the basis elements, then wedge them together the resulting number (1-d vector spaces are just numbers remember) is the volume of the object that the vectors span (we're normalizing things so that we consider the basis vectors of V to have length 1). It is easiest to justify this by saying: consider some box defined by n vectors, if you double all the lengths of all the vectors you multiply the volume by 2^n, which is exactly waht should happen to an n dimensional volume.
So, back to determinants. Given a linear map M acting on V I can make it act on /\^n(V) by just letting M act on each of the components of a wedge of n things at the same time. When n=dim(V) M now just acts by multiplication by a number. That number is what we define to be det(M). Using the properties of wedges you can also now work out a formula for the determinant of any nxn matrix, and it gives precisely the things you know in the 2x2 and 3x3 case. Here are the determinants written out for small matrices
M= m_{ij} for 1<=i,j<=2
det(M)=m_{11}m_{22} - m_{12}m_{21}[\itex]<br />
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now for i,j>=3<br />
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det(M)=<br />
m_{11}m_{22}m_{33}- m_{12}m_{21}m_{33}-m_{13}m_{22}m_{31} - m_{11}m_{23}m_{32} + m_{12}m_{23}m_{31}+m_{13}m_{21}m_{32}+m_{13}m_{22}m_{31}<br />
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I've deliberately written in that form so that you can try to spot a pattern: the subscripts all go m_{1i}m_{2j}m_{3k}[/tex] and {i,j,k}={1,2,3}, the sign in front tells me how many swaps I made in the order of 1,2,3 to get i,j,k, i.e. to get from 1,2,3 to 2,1,3 you need one swap, 1,2,3 to 2,3,1 needs two swaps. Obviously &#039;the number of swaps&#039; is not well defined, but what is well defined is the parity, no matter how you get from 1,2,3 to 2,1,3 you will always use an odd number, and from 1,2,3 to 2,3,1 an even number. <br />
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These formulae follow from the rules about things in /\^n. Try it for the 2x2 case and see what happens: If we take the usual 2x2 matrix with entries a,b,c,d, where does it send e_1? To ae_1+ce_2, and e_2 goes to be_1+de_2. Now what is<br />
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(ae_1+ce_2)/\(be_1+de_2)?<br />
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We can distribute over the brackets:<br />
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abe_1/\e_1 + ade_1/\e_2 + cbe_2/\e_1+cde_2/\e_2.<br />
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Remember repeated indices mean the symbol is equal to zero:<br />
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ade_1/\e_2+bce_2/\e_1<br />
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and we swap signs if we swap arguments:<br />
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ade_1/\e_2 - bc e_1/\e_2 = (ad-bc)e_1/\e_2<br />
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as required.