Solve the system of equations using matrices. Use Gaussian elimination with back-substitution.

Question

erinweeks · Answer

3x + 5y - 2w	=	-13 
2x + 7z - w	=	-1 
4y + 3z + 3w	=	1 
-x + 2y + 4z	=	-5

erinweeks · Answer

A. ( -1, 20/13, 0, 2/5 )
B. ( 1, -2, 0, 3 )
C. ( 3/4, -2, 0, 3/4 )
D. ( 4/3, - 13/20, 0, 5/2 )

ganeshie8 · Answer

can u write the equations in matrix form ?

erinweeks · Answer

|dw:1379709207452:dw|

erinweeks · Answer

i think?

ganeshie8 · Answer

careful, u have 4 variables, so u must have 4 clomns aswell

ganeshie8 · Answer

$$ 

\left[ \begin{array}{ccc}
3 & 5 & 0 & -2 \
2 & 0 & 7 & -1 \
0  & 4 & 3 & 3 \
-1 & 2 & 4 & 0 \
\end{array} \right]

\left[ \begin{array}{ccc}
-13 \
-1 \
1 \
-5 \
\end{array} \right]

$$

ganeshie8 · Answer

reduce it to triangular matrix

erinweeks · Answer

oh sorry ! im not good at this still lol

erinweeks · Answer

i dont know how to reduce?

ganeshie8 · Answer

R1-R2

ganeshie8 · Answer

subtract R2 from R1

ganeshie8 · Answer

$$ 

\left[ \begin{array}{ccc}
3 & 5 & 0 & -2 \
2 & 0 & 7 & -1 \
0  & 4 & 3 & 3 \
-1 & 2 & 4 & 0 \
\end{array} \right]

\left[ \begin{array}{ccc}
-13 \
-1 \
1 \
-5 \
\end{array} \right]

$$ 


R1-R2

$$ 

\left[ \begin{array}{ccc}
1 & 5 & -7 & -1 \
2 & 0 & 7 & -1 \
0  & 4 & 3 & 3 \
-1 & 2 & 4 & 0 \
\end{array} \right]

\left[ \begin{array}{ccc}
-12 \
-1 \
1 \
-5 \
\end{array} \right]

$$

ganeshie8 · Answer

ive just subtracted R2 from R1

ganeshie8 · Answer

see if u okays wid that

erinweeks · Answer

okay now what though?

ganeshie8 · Answer

next, we want to make first column as below :-
1
0
0
0

ganeshie8 · Answer

for that we do this :-
R2-2R1
R4+R1

erinweeks · Answer

now im confused can you show me lol!

ganeshie8 · Answer

$$ 

\left[ \begin{array}{ccc}
3 & 5 & 0 & -2 \
2 & 0 & 7 & -1 \
0  & 4 & 3 & 3 \
-1 & 2 & 4 & 0 \
\end{array} \right]

\left[ \begin{array}{ccc}
-13 \
-1 \
1 \
-5 \
\end{array} \right]

$$ 


R1-R2

$$ 

\left[ \begin{array}{ccc}
1 & 5 & -7 & -1 \
2 & 0 & 7 & -1 \
0  & 4 & 3 & 3 \
-1 & 2 & 4 & 0 \
\end{array} \right]

\left[ \begin{array}{ccc}
-12 \
-1 \
1 \
-5 \
\end{array} \right]

$$ 

R2-2R1
R4+R1

$$ 

\left[ \begin{array}{ccc}
1 & 5 & -7 & -1 \
0 & -10 & 21 & 0 \
0  & 4 & 3 & 3 \
0 & 7 & -3 & -1 \
\end{array} \right]

\left[ \begin{array}{ccc}
-12 \
23 \
1 \
-17 \
\end{array} \right]

$$

erinweeks · Answer

okay i get it

ganeshie8 · Answer

$$ 

\left[ \begin{array}{ccc}
3 & 5 & 0 & -2 \
2 & 0 & 7 & -1 \
0  & 4 & 3 & 3 \
-1 & 2 & 4 & 0 \
\end{array} \right]

\left[ \begin{array}{ccc}
-13 \
-1 \
1 \
-5 \
\end{array} \right]

$$ 


R1-R2

$$ 

\left[ \begin{array}{ccc}
1 & 5 & -7 & -1 \
2 & 0 & 7 & -1 \
0  & 4 & 3 & 3 \
-1 & 2 & 4 & 0 \
\end{array} \right]

\left[ \begin{array}{ccc}
-12 \
-1 \
1 \
-5 \
\end{array} \right]

$$ 

R2-2R1
R4+R1

$$ 

\left[ \begin{array}{ccc}
1 & 5 & -7 & -1 \
0 & -10 & 21 & 0 \
0  & 4 & 3 & 3 \
0 & 7 & -3 & -1 \
\end{array} \right]

\left[ \begin{array}{ccc}
-12 \
23 \
1 \
-17 \
\end{array} \right]

$$ 

R2/-10

$$ 

\left[ \begin{array}{ccc}
1 & 5 & -7 & -1 \
0 & 1 & -2.1 & 0 \
0  & 4 & 3 & 3 \
0 & 7 & -3 & -1 \
\end{array} \right]

\left[ \begin{array}{ccc}
-12 \
-2.3 \
1 \
-17 \
\end{array} \right]

$$

ganeshie8 · Answer

next, ive divided R2 by -10

erinweeks · Answer

why did you do divide by 10?

ganeshie8 · Answer

cuz i want 1 at 22 position (row2, col2)

erinweeks · Answer

okay gotcha!! just was confused for a sec

ganeshie8 · Answer

$$ 

\left[ \begin{array}{ccc}
3 & 5 & 0 & -2 \
2 & 0 & 7 & -1 \
0  & 4 & 3 & 3 \
-1 & 2 & 4 & 0 \
\end{array} \right]

\left[ \begin{array}{ccc}
-13 \
-1 \
1 \
-5 \
\end{array} \right]

$$ 


R1-R2

$$ 

\left[ \begin{array}{ccc}
1 & 5 & -7 & -1 \
2 & 0 & 7 & -1 \
0  & 4 & 3 & 3 \
-1 & 2 & 4 & 0 \
\end{array} \right]

\left[ \begin{array}{ccc}
-12 \
-1 \
1 \
-5 \
\end{array} \right]

$$ 

R2-2R1
R4+R1

$$ 

\left[ \begin{array}{ccc}
1 & 5 & -7 & -1 \
0 & -10 & 21 & 0 \
0  & 4 & 3 & 3 \
0 & 7 & -3 & -1 \
\end{array} \right]

\left[ \begin{array}{ccc}
-12 \
23 \
1 \
-17 \
\end{array} \right]

$$ 

R2/-10

$$ 

\left[ \begin{array}{ccc}
1 & 5 & -7 & -1 \
0 & 1 & -2.1 & 0 \
0  & 4 & 3 & 3 \
0 & 7 & -3 & -1 \
\end{array} \right]

\left[ \begin{array}{ccc}
-12 \
-2.3 \
1 \
-17 \
\end{array} \right]

$$ 


R3-4R2
R4-7R2

$$ 

\left[ \begin{array}{ccc}
1 & 5 & -7 & -1 \
0 & 1 & -2.1 & 0 \
0  & 0 & 11.4 & 3 \
0 & 0 & 11.7 & -1 \
\end{array} \right]

\left[ \begin{array}{ccc}
-12 \
-2.3 \
10.2 \
-0.9 \
\end{array} \right]

$$

ganeshie8 · Answer

plz do check if my calculations are correct or not :)

erinweeks · Answer

i believe they are!!! so is that it..

erinweeks · Answer

or what do we do from here?

ganeshie8 · Answer

no, we need to conver it upper triangulear

ganeshie8 · Answer

if u keep going, u should get the form below :- http://www.wolframalpha.com/input/?i=row+echleon++%7B%7B3%2C5%2C0%2C-2%2C-13%7D%2C%7B2%2C0%2C7%2C-1%2C-1%7D%2C%7B0%2C4%2C3%2C3%2C1%7D%2C%7B-1%2C2%2C4%2C0%2C-5%7D%7D

erinweeks · Answer

i didnt really get that.. that doesnt look like any of my answers

ganeshie8 · Answer

last column is the solution

erinweeks · Answer

that doesnt look like this though
A. ( -1, 20/13, 0, 2/5 )
B. ( 1, -2, 0, 3 )
C. ( 3/4, -2, 0, 3/4 )
D. ( 4/3, - 13/20, 0, 5/2 )

ganeshie8 · Answer

attachment

erinweeks · Answer

oh okay so its always the last column ?

ganeshie8 · Answer

yes