MHB 307.1.1 Use row reduction on the appropriate augmented

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The discussion focuses on using row reduction to solve a system of equations represented by an augmented matrix. The correct solutions found are x = 4/5 and y = -1/5, confirmed through substitution back into the original equations. Participants express frustration with the complexity of the row reduction process, describing it as "arithmetic torture," while others argue that it becomes easier with practice. The conversation also touches on the usefulness of calculators for handling such calculations. Overall, the thread highlights both the challenges and the methods involved in matrix row reduction.
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??
 
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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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