Row Reduction of a matrix

Question

Let us consider the following matrix $$M= \left[ {\begin{array}{cc} 1 & 1 & 2 & 1 & 5 \\ 1 & 1 & 2 & 6 & 10 \\ 1 & 2 & 5 & 2 & 7 \\ \end{array} } \right]$$

I was able to reduce the above matrix to a row echelon matrix:

$$M'=\left[ {\begin{array}{cc} 1 & 0 & -1 & 0 & 3 \\ 0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ \end{array} } \right]$$

But I don't know how to express M' as a multiplication by a sequence $E_1,...,E_k$ of elementary matrices: $$M'=E_k...E_2E_1M$$

Thank you in advance

Can you tell us the sequence of $k$ row operations that you used to reduce $M$ to $M'$? — Adriano, Aug 25 '13 at 05:07
@Adriano

$R_2 \rightarrow R_2 - R_1$

$R_3 \rightarrow R_3-R_1$

$ R_2 \leftrightarrow R_3$

$R_1 \rightarrow R_1- R_2$

$ R_3 \rightarrow \frac{1}{5} R_3$

$R_2 \rightarrow R_2 - R_3$ — amir, Aug 25 '13 at 05:13
Given that each row operation can be represented by a left-product with an invertible matrix $E$, can you find $E$ for the 3 types of row operation? — Jonathan Y., Aug 25 '13 at 05:16
@JonathanY. That is my question actually. Could you explain to me how to express each type of row operation as a elementary matrix ?
As in: 1) Linear combination $\Leftrightarrow$ elementary matrix 2) Multiplying a row by a scalar $\Leftrightarrow$ elementary matrix 3) Interchanging two rows $\Leftrightarrow$ elementary matrix — amir, Aug 25 '13 at 05:27

score 2 · Accepted Answer · answered Aug 25 '13 at 05:48

2

$M'=E_k...E_2E_1M$

From the row operations you provided, we arrive at:

$\begin{bmatrix} 1& 0& 0 \\0& 1& -1 \\ 0& 0& 1 \end{bmatrix} \begin{bmatrix} 1& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1/5 \end{bmatrix}\begin{bmatrix} 1& -1& 0 \\ 0& 1& 0 \\ 0& 0& 1 \end{bmatrix}\begin{bmatrix} 1& 0& 0 \\ 0& 0& 1 \\ 0&1& 0 \end{bmatrix}\begin{bmatrix} 1& 0& 0 \\ 0& 1& 0 \\ -1& 0& 1 \end{bmatrix}\begin{bmatrix} 1& 0& 0 \\ -1& 1& 0 \\ 0& 0& 1 \end{bmatrix}\begin{bmatrix} 1 & 1 & 2 & 1 & 5 \\ 1 & 1 & 2 & 6 & 10 \\ 1 & 2 & 5 & 2 & 7 \\ \end{bmatrix} = \begin{bmatrix}1 & 0 & -1 & 0 & 3 \\0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1\end{bmatrix}$

answered Aug 25 '13 at 05:48

Amzoti

56,093

$\ddot\smile$ thanks! – amWhy Aug 25 '13 at 14:19
No, actually, it was deleted because of the flags. I've flagged for moderator attention. Perhaps you can flag it for mod attention, too :-( – amWhy Aug 25 '13 at 20:47

Adriano · Answer 2 · 2013-08-25T05:46:26.447

1

Hint: To remember how to express each type of row operation as an elementary matrix, simply perform that row operation on the identity matrix. Thus, for example, the first elementary matrix $E_1$ corresponding to your first row operation (adding $(-1) \cdot R_1$ to $R_2$) would be: $$ E_1 = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

Can you see how this matrix was obtained? Can you see why its size is $3 \times 3$, and not $5 \times 5$?

edited Aug 25 '13 at 05:46

answered Aug 25 '13 at 05:38

Adriano

41,576

Thank you. I understand now. Have a nice day – amir Aug 25 '13 at 05:46

score 0 · Answer 3 · answered Aug 25 '13 at 05:39

If we take $A\in\mathbb{F}^{n\times m}$ and $x\in\mathbb{F}^n$, then $x^TA\in\mathbb{F}^m$ is a linear combination of $A$'s rows with coefficients $x_1,x_2\ldots,x_n$. It follows that for $E\in\mathbb{F}^{n\times n}$, each row of $EA$ is a linear combination of $A$'s rows with coefficients given by the corresponding row of $E$. This should be helpful in constructing the different types of elementary matrices.

Row Reduction of a matrix

3 Answers3