My Math Forum Matrix operations to row-echelon form - why?

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 July 18th, 2012, 12:30 AM #1 Member   Joined: Oct 2009 Posts: 50 Thanks: 0 Matrix operations to row-echelon form - why? Hey, I'm learning about matrices (without a teacher, I'm doing it by myself) And I've consulted different books, and I am consistently stumbling over a specific concept. When you're trying to "turn a matrix into row-echelon form", apparently there are three kinds of operations that you can use to do this. interchange two rows multiply a row by a non-zero scalar add to any row the elements of another row multiplied by any scalar. I understand these operations, but I just don't understand why this can just be done? If you do this, you are litteraly changing the matrix. It is no longer the same matrix. So what does it mean to "turn it into row-echelon form"? This implies that it is the same matrix, but in a different form? in simple arithmetic, when you have an equation: 3x +5 = 7x - 3. you can deduce from this: x = 2 by the rules of algebra. But I don't understand what the significance of the matrix operations is. Why can we just change the matrix like that? Thanks!
 July 18th, 2012, 05:53 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Matrix operations to row-echelon form - why? Every "row operation" corresponds to the "elementary matrix" that you get by performing that row operation on the identity matrix. For example, using 3 by 3 matrices for simplicity, the elementary matrix corresponding to "swap the first and third rows" is $\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$. The elementary matrix corresponding to "add 2 times the first row to the second row " is $\begin{bmatrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$. And the point is that applying a row operation to a matrix is the same as multiplying by the corresponding elementary matrix. So, to solve the equation AX= B where A, B, and X are matrices (which includes the case that X is a "column matrix", "vector"), We write A and B beside one another, then apply a series of row operations, $r_1$, $r_2$, ..., $r_n$ that reduce A to the identity matrix, at the same time, applying those row operations to B. In terms of elementary matrices, we are multiplying the corresponding elementary matrices, $R_1$, $R_2$, ..., $R_n$. Since $R_1R_2...R_nA= I$, $R_1R_2...Rn$ is the inverse matrix to A and, since applying those same row operations to B is the same as multiplying $R_1R_2...R_nB= A^{-1}B$ so that you wind up with $A^{-1}Ax= x= A^{-1}B$.
July 18th, 2012, 11:59 AM   #3
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Re: Matrix operations to row-echelon form - why?

Quote:
 Originally Posted by HallsofIvy Every "row operation" corresponds to the "elementary matrix" that you get by performing that row operation on the identity matrix. For example, using 3 by 3 matrices for simplicity, the elementary matrix corresponding to "swap the first and third rows" is $\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$. The elementary matrix corresponding to "add 2 times the first row to the second row " is $\begin{bmatrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$. And the point is that applying a row operation to a matrix is the same as multiplying by the corresponding elementary matrix. So, to solve the equation AX= B where A, B, and X are matrices (which includes the case that X is a "column matrix", "vector"), We write A and B beside one another, then apply a series of row operations, $r_1$, $r_2$, ..., $r_n$ that reduce A to the identity matrix, at the same time, applying those row operations to B. In terms of elementary matrices, we are multiplying the corresponding elementary matrices, $R_1$, $R_2$, ..., $R_n$. Since $R_1R_2...R_nA= I$, $R_1R_2...Rn$ is the inverse matrix to A and, since applying those same row operations to B is the same as multiplying $R_1R_2...R_nB= A^{-1}B$ so that you wind up with $A^{-1}Ax= x= A^{-1}B$.
So if I understand it correctly:
The situation in which you can apply the basic operations are when:
AX=B
, but then you MUST apply exactly the same changes to both A and B, am I correct?
In the same way that if you have:
5x = b, in algebra
that you can divide the left side by 5, but then you MUST also divide the right side by 5:
x = b/5
Am I correct, that this is conceptually the same thing, as the basic operators for matrices?
They are just rules for doing operations in matrix equations?

 August 16th, 2012, 08:03 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Matrix operations to row-echelon form - why? Yes, that is correct. All the "usual" rules of arithmetic apply except one- matrix multiplication is not commutative. That is, in general AB is not equal to BA. That's why we typically do not write "A/B"- that does not allow us to indicate an order. We write $A^{-1}B$ or $BA^{-1}$. If the equation is Ax= B, then to get x we must have $A^{-1}(Ax)$, not $(Ax)A^{-1}$. That is, $A^{-1}(Ax)= x= A^{-1}B$

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