307.1.1 Use row reduction on the appropriate augmented

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Discussion Overview

The discussion revolves around the process of using row reduction on an augmented matrix to solve a system of linear equations. Participants share their experiences, methods, and challenges encountered during the row reduction process.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a system of equations and their solution using row reduction, expressing hope that there are no typos in their work.
  • Another participant confirms the correctness of the solution by substituting the values back into the original equations.
  • Some participants express frustration with the complexity of the row reduction steps, describing the process as confusing and time-consuming.
  • One participant suggests that the difficulty lies in the arithmetic involved in matrix reduction, while another counters that it is primarily a matter of practice.
  • There is a mention of calculators being helpful for matrix reduction, with a participant sharing a link to a calculator image.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the solution provided, but there is a divergence in opinions regarding the difficulty of the row reduction process and the reliance on calculators.

Contextual Notes

Some participants express uncertainty about the clarity of the steps involved in row reduction, indicating that the arithmetic can be challenging and may require practice to master.

Who May Find This Useful

Students learning about linear algebra, particularly those struggling with row reduction techniques and matrix arithmetic.

karush
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$\tiny{307.1.1}$
Use row reduction on the appropriate augmented matrix to solve the following system of equations:
$\begin{array}{ll}3x+2y&=2\\x-y&=1\end{array}
\sim\left[\begin{array}{rr|r}3&2&2\\ \:1&-1&1\end{array}\right]\sim
\begin{bmatrix}1&0&\frac{4}{5}\\ 0&1&-\frac{1}{5}\end{bmatrix}$

$x=\dfrac{4}{5}\quad y= -\dfrac{1}{5}$

hopefully no typos!
suggestions??
 
Last edited:
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Well, 3(4/5)+ 2(-1/5)= 12/5- 2/5= 10/5= 2
and 4/5- (-1/5)= 4/5+ 1/5= 5/5= 1

So you certainly got the right answers!
 
as simple as it looks the steps were very confusing
spent over an hour on it😕
 
karush said:
as simple as it looks the steps were very confusing
spent over an hour on it😕

reduction of matrices is arithmetic torture ...that's why calculators were invented

matrix_red1.png

matrix_red2.png
 
I don't consider it that bad. It's mostly just a matter of practice like arithmetic.
\[ \begin{array}{c}-3\\*\end{array}\left[\begin{array}{cc|c}3&2&2\\1&-1&1\end{array}\right] \to \updownarrow\left[\begin{array}{cc|c}0&5&-1\\1&-1&1\end{array}\right] \to \begin{array}{c}\\*\frac 15\end{array}\left[\begin{array}{cc|c}1&-1&1\\0&5&-1\end{array}\right] \to \begin{array}{c}+1\\*\end{array}\left[\begin{array}{cc|c}1&-1&1\\0&1&-\frac 15\end{array}\right] \to \left[\begin{array}{cc|c}1&0&\frac 45\\0&1&-\frac 15\end{array}\right] \]
 
skeeter said:
reduction of matrices is arithmetic torture ...that's why calculators were invented

https://www.physicsforums.com/attachments/10876
View attachment 10877
\looks like a TI

I had a a TI CAS but loaned it out never got it back
 

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