307.1.1 Use row reduction on the appropriate augmented

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SUMMARY

The discussion focuses on solving a system of equations using row reduction on an augmented matrix. The specific equations presented are 3x + 2y = 2 and x - y = 1, which are represented in augmented matrix form as [3 2 | 2; 1 -1 | 1]. The final solutions derived from the row reduction process are x = 4/5 and y = -1/5. Participants express that while the process appears simple, it can be confusing and time-consuming, emphasizing the importance of practice in mastering matrix reduction techniques.

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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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