
Elementary Matrices...
I really need help with this question...
I need to find the elementary matrice so that:
EB=C
B = $\displaystyle \[ \left( \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 0 & 1 \\ 0 & 6 & 6 \end{array} \right)\]$
C = $\displaystyle \[ \left( \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{array} \right)\]$
This is what i've tried
Row 3 < 1/6R3 (Doesn't work...)
Any other methods don't work since they add different values..
Can someone help me out here please?
I've spent an hour on this question already (Worried)

You've got the right elementary row operation. What you really need to think about is which elementary matrix has that same effect?

Well the elementary matrix I got was:
E = $\displaystyle \[ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1/6 \end{array} \right)\]$
and i'm pretty sure that's wrong since there has to be all leading ones going down diagonally...

You're thinking of the identity matrix, I think. Look here, and you'll see that an elementary matrix is obtained from the identity matrix by performing one elementary row operation on it.

An elementary matrix is simply a matrix derived from the identity matrix by a single rowoperation.
There are three types of row operations: add a multiple of one row to another, swap one row with another, and multiply one row by a number.
For the first of those, adding a multiple of one row to another, yes, the diagonal remains "1"s. But for the other two that is not true. The matrix you give is an elementary matrix and is the correct answer to this problem. You go from B to C by the row operation "multiply the third row by 1/6" so you get the corresponding elementary matrix by multiplying the third row of the identity matrix by 1/6, just as you have.

Thanks guys, I understand this now.

You're welcome for whatever I contributed. Have a good one!