Linear algebra EA=R factorization question

[FocusedWolf]

ok i know about the thing, i think its called gauss-jordan, where you do elimination on [A I] and you get [I A^-1] or [R E]or something like that.

question is how can you get the E that puts A into reduced row echelon form with "row operations" only...supposedly some easy way ?

on pg 134 Gilbert Strangs book "introduction to linear algebra" he says:

"EA = R
The square matrix E is the product of elementary matrices Eij and Pij and D^-1..." yea that's all he says about doing that, but it was good enough to get it on the exam, and probably the final now lol

on pg 139 he does some work that doesn't look like steps but he gets two different E that can get A unto rref... so what's the secret? :P

 

[Hurkyl]

I'm a little confused; what exactly is it that you want to know?

 

[HallsofIvy]

ok i know about the thing, i think its called gauss-jordan, where you do elimination on [A I] and you get [I A^-1] or [R E]or something like that.

question is how can you get the E that puts A into reduced row echelon form with "row operations" only...supposedly some easy way ?

on pg 134 Gilbert Strangs book "introduction to linear algebra" he says:

"EA = R
The square matrix E is the product of elementary matrices Eij and Pij and D^-1..." yea that's all he says about doing that, but it was good enough to get it on the exam, and probably the final now lol

on pg 139 he does some work that doesn't look like steps but he gets two different E that can get A unto rref... so what's the secret? :P

Well, yes, there is an "easy way" to get E- you DO the row operations that put A into "reduced echelon form". For example, to find the inverse of
$$A= \left[\begin{array}{cc}2 & 1 \\ 3 & 2\end{array}\right]$$
First set it up as:
$$\left[\begin{array}{cccc}2 & 1 & 1 & 0\\ 3 & 2 & 0 & 1\end{array}\right]$$
Now do row operations to reduce the "A" part to the identity matrix (ie.e reduced row echelon). Do one column at a time. Since you want a "1" in the upper left, divide the first row by 2. Since you want a "0" in the lower left, multiply the new first row by -3 and add to the second row:
$$\left[\begin{array}{cc}1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & -\frac{3}{2} & 1\end{array}\right]$$
Now you want the second column to be "0" and "1": multiply the second row by 2 and subtract 1/1 times the new second row from the first row:
$$\left[\begin{array}{cc}1 & 0 & 2 & -1 \\ 0 & 1 & -3 & 2\end{array}\right]$$
The inverse matrix is
$$A^{-1}= \left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$$

The quote from Strang is not saying HOW to do that but explaining WHY it works. Every row operation corresponds to an elementary matrix constructed by applying that row operation to the identity matrix: applying the row operation to any matrix is the same as multiplying the matrix by the corresponding elementary matrix. Applying a sequence of row operations is the same as multiplying by the corresponding elementary matrices in order. The product of all those elementary matrices is the same as applying all the row operations to the identity matrix.

In the above calculation, the sequence of row operations reduces A to the identity matrix so multiplying by the product of their corresponding elementary matrices does the same: that product is the inverse matrix. Since we apply those same operations to the identity matrix as we work, we produce that inverse matrix.