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[SOLVED] 307.1.1 Use row reduction on the appropriate augmented

karush

Well-known member
Jan 31, 2012
3,066
$\tiny{307.1.1}$
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:

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
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!
 

karush

Well-known member
Jan 31, 2012
3,066
as simple as it looks the steps were very confusing
spent over an hour on it😕
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
1,005
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
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
I don't consider it that bad. It's mostly just a matter of practice like arithmetic.
\[ \begin{array}{c}-3\\*\end{array}\left[\begin{array}{cc|c}3&2&2\\1&-1&1\end{array}\right] \to \updownarrow\left[\begin{array}{cc|c}0&5&-1\\1&-1&1\end{array}\right] \to \begin{array}{c}\\*\frac 15\end{array}\left[\begin{array}{cc|c}1&-1&1\\0&5&-1\end{array}\right] \to \begin{array}{c}+1\\*\end{array}\left[\begin{array}{cc|c}1&-1&1\\0&1&-\frac 15\end{array}\right] \to \left[\begin{array}{cc|c}1&0&\frac 45\\0&1&-\frac 15\end{array}\right] \]
 

karush

Well-known member
Jan 31, 2012
3,066