How to get this matrix to RREF

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July 6th 2010, 07:24 PM
Jasked

How to get this matrix to RREF

I'm just trying to figure out how to get this to rref:
x + 5y + z = 0
x + 6y - z = 0
2x + ay + bz = c

1 5 1 0
1 6 -1 0
2 a b c

I know the following is the answer, but I'm now sure how to get there. The problem I'm having is that I don't know how to handle the a, b and c. I probably just need someone to explain how to do the bottom row.
1 0 0 -11c/(2a+b-22)
0 1 0 2c/(2a+b-22)
0 0 1 c/(2a+b-22)

Thanks
July 6th 2010, 07:38 PM
dwsmith

$\begin{bmatrix} 1 & 5 & 1 & 0\\ 1 & 6 & -1 & 0\\ 2 & a & b & c \end{bmatrix}$

You want this form:
$\begin{bmatrix} 1 & 0 & 0 & x\\ 0 & 1 & 0 & y\\ 0 & 0 & 1 & z \end{bmatrix} \ \mbox{where} \ x,y,x \in\mathbb{Z}$

Subtract row2 from row1; subtract 2*row1from row3

$\begin{bmatrix} 1 & 5 & -1 & 0\\ 0 & 1 & -2 & 0\\ 0 & 10-a & -2-b & -c \end{bmatrix}$

Next subtract -5*row from row1, etc, etc, until you arrive at, or close to:

$\begin{bmatrix} 1 & 0 & 0 & x\\ 0 & 1 & 0 & y\\ 0 & 0 & 1 & z \end{bmatrix} \ \mbox{where} \ x,y,x\in\mathbb{Z}$
July 6th 2010, 11:35 PM
Jasked

That's taking me up to about the point where I don't know what to do.
For example, I'm not sure how the bottom row goes from this:
2 a b c
to this:
0 0 1 c/(2a+b-22)

If there were no a, b and c, I wouldn't have any problems with this question. I understand the basic row operations, but if you throw a, b and c in there then I'm not sure what to do. The basic row operations don't appear to be enough to end up with c/(2a+b-22) on the right hand side of the bottom. Pardon me if this isn't technically called RREF (I'm not sure), but that's the solution I'm trying to work towards.
July 7th 2010, 12:05 AM
Opalg

Quote:

Originally Posted by Jasked

That's taking me up to about the point where I don't know what to do.
For example, I'm not sure how the bottom row goes from this:
2 a b c
to this:
0 0 1 c/(2a+b-22)

If there were no a, b and c, I wouldn't have any problems with this question. I understand the basic row operations, but if you throw a, b and c in there then I'm not sure what to do. The basic row operations don't appear to be enough to end up with c/(2a+b-22) on the right hand side of the bottom. Pardon me if this isn't technically called RREF (I'm not sure), but that's the solution I'm trying to work towards.

Having got to the matrix $\begin{bmatrix} 1 & 5 & 1 & 0\\ 0 & 1 & -2 & 0\\ 0 & a-10 & b-2 & c \end{bmatrix}$ , the next step is to subtract $(a-10)$ times row 2 from row 3. That gives you the matrix

$\begin{bmatrix} 1 & 5 & 1 & 0\\ 0 & 1 & -2 & 0\\ 0 & 0 & x & c \end{bmatrix}$ , where $x = (b-2) - (a-10)(-2) = 2a+b-22$ . For the following step, divide the new row 3 by $2a+b-22$ .
July 7th 2010, 11:00 AM
Jasked

Opalg, that was exactly what I needed to read. Thank you so much for your help.