Collinearity, coplanarity and determinant

I don't get what the question is asking me. I'm confused why they add '1's to the matrix. Anyway here's my attempt.

For part (i), my analysis is that since P2, P3, P4 are not collinear, therefore they are linearly independent. This is to say that the last 3 rows in the matrix is linearly independent. Since det(matrix) = 0. The first row of the matrix must be a linear combination of the last 3 rows. Hence det(matrix) gives zero, which is the volume of a plane in R^4.

For part(ii), denote the matrix as A. Since P1, P2, P3, P4 are coplanar, they are linearly dependent. Hence det(A) = 0.

Now I try to show the converse. Since, det(A) = 0, Ax=0 has non-trivial solutions. Hence the set {P1,P2,P3,P4} are linearly dependent. Hence they are coplanar.

This is as far as I can go. Hope you guys can provide me with guidelines to steer me on the right track.

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They add $1$'s because it is an affine question. If it was a vector question, you would have one point less and no $1$'s. –  1015 Mar 18 '13 at 11:01
Seems like you answered the questions correctly. –  yohBS Mar 18 '13 at 11:11
Oh ok. I don't know about affine. This is a basic linear algebra course I'm taking. So what should I do when dealing with affine questions? –  uohzxela Mar 18 '13 at 11:15
@yohBS: Really? I thought my answers need more mathematical rigor.. –  uohzxela Mar 18 '13 at 11:18
It is not quite true that if $P_2,P_3,P_4$ are not collinear than they are linearly independent; take for instance $(1,0,0)$, $(0,1,0)$ and $(1,1,0)$ which are neither collinear nor independent. What is true (but not proved above) is that $3$ non-collinear points give linearly independent vectors after tacking on a coordinate $1$ to the end of each of them. –  Marc van Leeuwen Mar 18 '13 at 11:21