What we end up with is the row echelon form of the system.
Note: That we are using row 1 to change column 1 and row 2 to change column 2, in that order, so that we don't lose the 0's that we form.
Now we come to the Back Substituion stage. We first solve for the varible in the last row, then use that result to solve for the second last variable and so on.
[tex]\begin{alignedat}{1}\frac{9}{2}z & =\frac{15}{2}
\\ \therefore z & =\frac{5}{3}
\\ y+\frac{3}{2}z & =\frac{3}{2}
\\ y & =\frac{3}{2}-\frac{3}{2}\left(\frac{5}{3}\right)
\\ & =\frac{3}{2}-\frac{5}{2}
\\ \therefore y & =-1
\\ 4z+2y-z & =1
We may also sometimes use euler's identity ...
I just also wanna mention that the cross product is only defined for vectors that are in R^3.
Consider three vectors a, b, c in R^3
define the scalar triple product to be a. (b x c)
then the volume of a parallelepiped determined by a, b, c is given by |a.(bxc)|
yeah ok it's not that hard, im sure you can see it from the diagram, that is how i derive it
V = Ah = |axb| |c||cos(phi)| = |c.(axb)|
also the scalar triple product is a cool way of showing that 3 vectors are coplanar. ie a.(bxc) = 0 then a, b, c are coplanar (why?)
(note: the use of straight lines on the matrices is a shorthand for the determinant -- you may not have seen it before, I hadn't seen it before uni...)
So we're left with a few (2 x 2) determinants, which we know how to find. But what exactly did I do there? Well, the , and came from the first row. The matrix determinant came from temporarily 'deleting' the row and column that came from:
And similarly, the matrix determinant that is multiplied by came from 'deleting' the row and column that is in:
But there's one more thing: the negative sign in front of the . Whether the coefficient of each element is 1 or -1 can be determined from the following representation:
So basically, starting at the first element (first row, first column), it's positive, then it alternates between positive and negative. This can be extended to higher dimensions, e.g. for a (6 x 6):
So, for example, if we expanded the second row instead, we would have:
If you expand this out you'll find it's exactly the same as the other expression.