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avatar+141 

i have tried over 6 times and ive been stuck on it for the past hour, can someone please explain how this can be solved? 

\(\begin{bmatrix} 1 & -1&-2 |-6\\ 3 & 2&0 |25\\ -4&1&-1 |12 \end{bmatrix}\)


sorry if this is not very legible, here is the equation for it, i just tried to convert it into an augmented matrix :')

\(x-y-2z=-6 \)

\(3x+2y=-25\)

\(-4x+y-z=12\) 


Thank you so much. 

edit: i have been trying for the past 3 hours and i still cannot solve this lol 

 Sep 11, 2019
edited by Nirvana  Sep 11, 2019
 #1
avatar+5776 
+1

well for one you've got 25 in the matrix where it should be -25.

 

\(\begin{pmatrix} 1&-1&-2&|&-6\\ 3&2&0&|&-25\\ -4&1&-1&|&12 \end{pmatrix}\\ \begin{pmatrix} 1&-1&-2&|&-6\\ 0&5&6&|&-7\\ -4&1&-1&|&12 \end{pmatrix}\\ \begin{pmatrix} 1&-1&-2&|&-6\\ 0&5&6&|&-7\\ 0&-3&-9&|&-12 \end{pmatrix}\\\)

 

Now exchange rows 2 and 3 and normalize row 2

\(\begin{pmatrix} 1 & -1 & -2 & -6 \\ 0 & 1 & 3 & 4 \\ 0 & 5 & 6 & -7 \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & 3 & 4 \\ 0 & 5 & 6 & -7 \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & -9 & -27 \end{pmatrix}\\ \)

 

Normalize row 3 and continue

 

\(\begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\\ \)

 

\(\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-5\\-5\\3\end{pmatrix}\)

.
 
 Sep 11, 2019
 #2
avatar+141 
0

oh my god is that why i kept getting the wrong answer,,, thank you for pointing my mistake out!! :) 

 
Nirvana  Sep 12, 2019

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