Question

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

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

Answer 1

Anyone help me solve this one please

Answer 2

Ok, first write down the system in augmented form AX=B where A are the coefficients of the system on d LHS, B is the vector on the RHS and X are the valiables. For example,

Answer 3

3 5 -2 -13
2 7 -1 -1
4 3 3 1
-1 2 4 -5

Answer 4

@da_ScienceMan I didnt understand how to do it. Could you please explain?

Answer 5

Yes very well then....u almost got the matrix right. First let us correct the matrix

Answer 6

ok

Answer 7

then what is the second step @da_ScienceMan

Answer 8

\[\left[\begin{matrix}3 & 5 & 0 & -2& | -13\\ 2& 0 & 7&-1 &|-1 \\0 & 4 & 3& 3 &| 1\\-1 & 2 & 4 & 0 & | -5\end{matrix}\right]\]

Answer 9

this is the matrix A and then the vector is given by \[\left(\begin{matrix}x \\ y \\ z \\ w\end{matrix}\right)\]

Answer 10

The next step is to reduce the matrix by guessing numbers such that u have almost all elements of the last rows equal to zero-

Answer 11

Next, U proceed and have the second to the last row elements also zero but with 1 fewer than the last row....its a long process but u can do it.

Answer 12

can you show me how to do it fro here. I am stuck over here exactly

Answer 13

Yes sure hang on a min!

Answer 14

i can tell u what to do step by step whil u write it down r u ready?

Answer 15

first, I will divide row 2 by 2, next I multiply row 2 by 1 and add it to row 4. Then I have a new matrix.....

Answer 16

Yes i am ready

Answer 17

\left[\begin{matrix}3 & 5 & 0 & -2 & | -13 \\ 2 & 0 & 7 & -1 & | -1 \\0 & 4 & 3 & 3 & | +1 \\ -1 & 2 & 4 & 0 & | -5 \\\end{matrix}\right]rightarrow I will beginy by swapping rows 3 and row 4 to get the following matrix \[\left[\begin{matrix}3 & 5 & 0 & -2 & | -13 \\ 2 & 0 & 7 & -1 & | -1 \\-1 & 2 & 4 & 0 & | -5 \\ 0 & 4 & 3 & 3 & | 1 \\\end{matrix}\right]\] rightarrow

Answer 18

i am trying to type the matrices after each step but it seems i made a mistake. Please swap row 3 and row 4

Answer 19

Next divide row 2 by 2

Answer 20

ok wait a minute

Answer 21

\[\left[\begin{matrix}3 & 5 & 0 & -2 & | -13 \\ 2 & 0 & 7 & -1 & | -1 \\0 & 4 & 3 & 3 & | +1 \\ -1 & 2 & 4 & 0 & | -5 \\\end{matrix}\right]\rightarrow \left[\begin{matrix}3 & 5 & 0 & -2 & | -13 \\ 2 & 0 & 7 & -1 & | -1 \\-1 & 2 & 4 & 0 & | -5 \\ 0 & 4 & 3 & 3 & | 1 \\\end{matrix}\right]\]

Answer 22

3 5 0 -2 | -13
1 0 3.5 -0.5 |-0.5
0 4 3 3 | 1
-1 2 4 0 |-5

Answer 23

@da_ScienceMan

Answer 24

then next step we get that \[\rightarrow \left[\begin{matrix}3 & 5 & 0 & -2 & | -13 \\ 1 & 0 & \frac{ 7 }{ 2 } & \frac{ -1 }{ 2} & | \frac{ -1 }{ 2 } \\-1 & 2 & 4 & 0 & | -5 \\ 0 & 4 & 3 & 3 & | 1 \\\end{matrix}\right]\]

Answer 25

right tight. Next, U multiply row 2 by 1 and add it row 3!

Answer 26

3 5 0 -2 |-13
1 0 7/2 -1/2 |-1/2
0 2 15/2 -1 | -11/2
0 4 3 3 | 1
@da_ScienceMan

Answer 27

Very good. Next task: multiply row 3 by 2 and add it to row lets see what u get...remember our aim is to have the three first terms in row 4 all zero, first two terms in row 3 all zero and 1st term in row zero!

Answer 28

multipliy row 3 by -2 and add it to row 4 i mean sorry!

Answer 29

ok

Answer 30

check the third term in row three in the above matrix u wrote and make SURE it is correct!

Answer 31

3 5 0 -2 |-13
1 0 7/2 -1/2 |-1/2
0 -4 -15 2 | 11
0 0 -12 5 |12

Answer 32

@da_ScienceMan is this correct

Answer 33

ok I wrote the steps using my rough sheet and m sure u or me is wrong. Now let us go through it again if u dont mind...

Answer 34

ok

Answer 35

i made a mistake in the third row fourth term

Answer 36

First u have to swap rows 3 and 4...which m sure u did. Next u multiply row 2 by 1/2 which u did.

Answer 37

Sure u can just use my last matrix and proceed from there......I think it is correct! pls check properly its been a while I did this thing.

Answer 38

ok wait let me write it down again

Answer 39

Yes that would be nice...first swap rows3 and 4. Next multiply row 2 by 0.5. Then multiply row 2 by 1 and add to row 3. Next, multiply row 3 by -2 and add to row 4. Our aim is to have first three terms all zero in row 4, first two terms all zero in row and first term zero in row 3. Then we can proceed from there.

Answer 40

@da_ScienceMan This is what i got

3 5 0 -2 |-13
1 0 7/2 -1/2 |-1/2
0 -4 -15 1 |-9
0 0 -12 4 |-8

Answer 41

i didnt get that....d first two rows seem correct to me. but the last two???

Answer 42

hang on a min. u just have to add d result u got from multiplying row 3 by adding them to row 4. keep row 3 as it is for nw!!!

Answer 43

can you do it because i m confused

Answer 44

so if u do it that way, u ll have [0 2 15/2 -1/2 | -11/2] for row 3, and [0 0 -12 4| 12] for row 4. pls verify!

Answer 45

my last metrix is correct

Answer 46

\[\left[\begin{matrix}3 & 5 & 0 & -2 & | -13 \\ 1 & 0 & \frac{ 7 }{ 2 } & \frac{ -1 }{2 } & \frac{ -1 }{ 2 } \\ 0 & 2 & \frac{ 15 }{ 2 } & \frac{ -1 }{2 } &| \frac{ -11 }{ 2 } \\ 0 & 0 & -12 & 4 &| 10\end{matrix}\right]\]

Answer 47

is d expected matrix

Answer 48

pls u can multiply row 3 and row 2 by 2 to make it easier for u to proceed!

Answer 49

i am realy frustrated .. i will go out for dinner i will calculate later... can you just write down the steps and i will try solving it.. if i didnt understand we will continue later if you online

Answer 50

@da_ScienceMan

Answer 51

ok sure its just easy for sure. it is just because m kinda laid back typing d matrices. no worries if coming back i can write down d steps, nd makes them available too. Just relax nd go for dinner. I ll be on for sometime!

Answer 52

OK THNXS ALOT @da_ScienceMan

Answer 53

sure no worries!

Answer 54

please check the steps in d attached file and verify. I represented d steps using R for rows together with the corresponding operation....

Answer 55

I claim that x is 6.6, y is -1.4, z is 6.3 and w is 1.1. But please verify!