Define matrix to get a row operation of type 1

  • Context: MHB 
  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Matrix Row Type
Click For Summary
SUMMARY

The discussion focuses on defining a matrix \( b \in \mathbb{R}^{n \times n} \) that facilitates a row operation of type 1, specifically multiplying one row by a scalar and adding it to another row. The matrices involved are \( a = \begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix} \), \( E_{1,3} = \begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix} \), and \( u_n = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} \). The operation is exemplified by the calculation \( (u_n + 2E_{1,3})a = \begin{pmatrix}15 & 18 & 21 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix} \), demonstrating how to apply the defined matrix to achieve the desired row operation.

PREREQUISITES
  • Understanding of matrix operations and linear algebra
  • Familiarity with row operations in matrix theory
  • Knowledge of matrix notation and manipulation
  • Basic concepts of scalar multiplication in matrices
NEXT STEPS
  • Study the properties of elementary matrices in linear algebra
  • Learn about different types of row operations and their applications
  • Explore the concept of matrix transformations and their geometric interpretations
  • Investigate the role of row echelon form in solving linear systems
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, matrix theory, or anyone involved in computational mathematics and algorithm development.

mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

We have the matrices \begin{equation*}a=\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix}, \ \ E_{1,3}=\begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}, \ \ u_n=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 &0 & 1\end{pmatrix}\end{equation*}

I have calculated:
\begin{align*}\left (u_n+2E_{1,3}\right )a&=\left (\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 &0 & 1\end{pmatrix}+2\begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\right )\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix} \\ &=\left (\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 &0 & 1\end{pmatrix}+\begin{pmatrix}0 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\right )\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix} \\ &= \begin{pmatrix}1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 &0 & 1\end{pmatrix}\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix} \\ &= \begin{pmatrix}15 & 18 & 21 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix}\end{align*}

Let $a \in \mathbb{R}^{n\times m}$. Determine a matrix $b \in \mathbb{R}^{n\times n}$, such that the product $ba$ is a row operation of type 1.

The row operatioon of type 1 is that we multiply one row by a scalar and add it to another row. For that do we define $b$ to be as above, i.e. in the form $\left (u_n+sE_{i,j}\right )$ ? (Wondering)
 
Physics news on Phys.org
Hey mathmari!

Yep.
In the example we saw that we added 2 times row 3 to the first row.
So now we generalize to s times row j that we add to row i. (Thinking)
 
Klaas van Aarsen said:
Yep.
In the example we saw that we added 2 times row 3 to the first row.
So now we generalize to s times row j that we add to row i. (Thinking)

Ok! Thank you! (Sun)
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 34 ·
2
Replies
34
Views
2K
Replies
31
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K