Question

Hey everybody... Please explain my question ...

Answer 1

It's now or never :P

Answer 2

heehehe wait just my pics are uploading... :)

Answer 3

Doesn't this belong in Physics? D:

Answer 4

this is the question of circuit analysis by david irwin...I want u guys to tell me how to solve these equations by matrices...!

Answer 5

ohh.. must have spoken too soon D:

Answer 6

no dear... read my complete question...and plz help me out...these two pages is the complete question... This is the question of mathematics... how to solve 4 or 5 equations by matrices?

Answer 7

u understand my question @terenzreignz ?

Answer 8

It's bad enough I don't know this Physics stuff, but matrices aren't my thing... let's have some more eyes on it ^_^
@ganeshie8

Answer 9

Although, when it comes to systems, there might be something I can scrape up...

Answer 10

okey let it be then...!
I want u guyz to explain me the method of solving 4 equations by matrices... not by simultaneously solving them...

Answer 11

I hope ur reply might not get too late :)

Answer 12

Okay, let's see...
\[\Large \left.\begin{matrix}159v_2-43v_3-12.6v_4+12.6v_5=756\\v_2-2v_4-v_5=-24\\-v_2+v_3-12v_4+12v_5=45\\5v_4-4v_5=12\end{matrix}\right.\]

Answer 13

Everything correct?

Answer 14

@waleed_imtiaz

Answer 15

here in the first equation.... its not 12.6 ...its 126... everything else is correct/..

Answer 16

\[\Large \left.\begin{matrix}159v_2-43v_3-126v_4+126v_5=756\\v_2-2v_4-v_5=-24\\-v_2+v_3-12v_4+12v_5=45\\5v_4-4v_5=12\end{matrix}\right.\]

Answer 17

okey thats fine... now how to solve by matrices...?

Answer 18

Okay then, first, make a matrix containing all the coefficients, in proper order, for v2, v3, v4, and v5.
\[\Large \left[\begin{matrix}159&-43&-126&126\\1&0&-2&-1\\-1&1&-12&12\\0&0&5&-4\end{matrix}\right]\]

Answer 19

okey got it... then next ???

Answer 20

It's multiplied to this column matrix:
\[\Large \left[\begin{matrix}159&-43&-126&126\\1&0&-2&-1\\-1&1&-12&12\\0&0&5&-4\end{matrix}\right]\left[\begin{matrix}v_2\\v_3\\v_4\\v_5\end{matrix}\right]\]

Answer 21

got it... I knoww... and it will equals the matrix of all hte coefficients in the equation after the equal sign.... then next?

Answer 22

\[\Large \left[\begin{matrix}159&-43&-126&126\\1&0&-2&-1\\-1&1&-12&12\\0&0&5&-4\end{matrix}\right]\left[\begin{matrix}v_2\\v_3\\v_4\\v_5\end{matrix}\right]=\left[\begin{matrix}756\\-24\\45\\12\end{matrix}\right]\]

yeah...

Answer 23

Did your instructor do an example problem? How did your instructor do it, with row-operations?

Answer 24

thats okey.... thne?

Answer 25

Kramer's rule? or perhaps inverses?

Answer 26

I need help :(
@hartnn ?

Answer 27

I dont know how to solve it and my professor also didn't tell me because the time was short.... so I consider it a few methods of solving it...I think it can be done by row or column operations? or by crammer rule... or if i go with inverses,,, it will be a big question .... its bit of a mess.... taking inverse of 4 by 4 matrix... is a big task... so i think we should go with the method that uses less mathematical calculations...

Answer 28

yea i need help too :( @hartnn @amistre64

Answer 29

reduce it to triangular form and back substitute

Answer 30

can u solve it for me? :(

Answer 31

i want complete solution because i tried it but failed... reducing it to upper or lower triangular matrix ...is a mess....

Answer 32

i dont see why its a mess yet, let me try :)

Answer 33

yep ! wolfram is giving all fractions, so manually solving using matrices is pain

http://www.wolframalpha.com/input/?i=solve++%7B%7B159%2C-53%2C-126%2C126%7D+%2C+%7B1%2C0%2C-2%2C-1%7D%2C+%7B-1%2C1%2C-12%2C12%7D%2C+%7B0%2C0%2C5%2C-4%7D%7D%7B%7Bw%7D%2C%7Bx%7D%2C%7By%7D%2C%7Bz%7D%7D+%3D+%7B%7B756%7D%2C+%7B-24+%7D%2C+%7B45%7D%2C+%7B12%7D%7D

Answer 34

yes...! just show me how u can solve it...? by taking inverses or by crammers rule or by row or column operations... @ganeshie8

Answer 35

okay ! lets do elimination

Answer 36

first exchange R1 & R2, i hope you're familiar wid row operations a bit ?

Answer 37

but u can tell me :) because its long time when i studied these matrices... thats why I am confused... may be i guess that when we interchange the rows or columns we take negative sign outside? right?

Answer 38

thats for determinants,
but when doing row operations, we can interchange rows freely (think of each row as an equaiton)
order of equations doesnt matter

Answer 39

okey then interchanging rows 1 and 2... then?

Answer 40

can we reduce it to upper or lower triangular matrix? my frnd told me to do so...and then solve the augmented matrix...

Answer 41

augmented matrix :-

159 -43 -126 126 756
1 0 -2 -1 -24
-1 1 -12 12 45
0 0 5 -4 12

Answer 42

our goal is to change it to upper triangular matrix

Answer 43

yes okey.... then how we'll change it to upper triangular matrix?

Answer 44

exchange R1 & R2


1 0 -2 -1 -24
159 -43 -126 126 756
-1 1 -12 12 45
0 0 5 -4 12

Answer 45

next do, R2 - 159R1

Answer 46

and, R3+R1

Answer 47

Are u going to make the first column asn this?
1
0
0
0

Answer 48

@ganeshie8

Answer 49

Yes,


1 0 -2 -1 -24
0 -43 192 -33 4572
0 1 -14 11 21
0 0 5 -4 12

Answer 50

you pls verify my calculations also ok :)

Answer 51

Next, exchange R2 & R3

Answer 52

1 0 -2 -1 -24
0 1 -14 11 21
0 -43 192 -33 4572
0 0 5 -4 12

Answer 53

R3 + 43R2

Answer 54

wait wait wait...how u wrote -33 on the second row wihile doing R2-159R1?

Answer 55

waiting for ur reply...!!!

Answer 56

yes thats a mistake, just checking if you're alert or not ;)


1 0 -2 -1 -24
0 -43 192 285 4572
0 1 -14 11 21
0 0 5 -4 12

Answer 57

oh hoo :p thats good :)

Answer 58

does that look correct right after doing, R2-159R1

Answer 59

yes...

Answer 60

ok lets keep going

Answer 61

Next, exhange R2 & R3

Answer 62

1 0 -2 -1 -24
0 1 -14 11 21
0 -43 192 285 4572
0 0 5 -4 12

Answer 63

R3+43R1

Answer 64

1 0 -2 -1 -24
0 1 -14 11 21
0 0 -410 758 5475
0 0 5 -4 12

Answer 65

exchange R3 & R4

Answer 66

1 0 -2 -1 -24
0 1 -14 11 21
0 0 5 -4 12
0 0 -410 758 5475

Answer 67

R3/5

Answer 68

1 0 -2 -1 -24
0 1 -14 11 21
0 0 1 -4/5 12/5
0 0 -410 758 5475

Answer 69

i have to go to MASJID...can we talk about this at nite? or u give me the solution and i will see it when I will come online...okey :) I am understanding ur calculations...

Answer 70

R4+410R3

Answer 71

ok np :) u go, il complete

Answer 72

okey I will see after I come back... :) thanks @ganeshie8

Answer 73

1 0 -2 -1 -24
0 1 -14 11 21
0 0 1 -4/5 12/5
0 0 0 430 6459

Answer 74

R4/430

Answer 75

1 0 -2 -1 -24
0 1 -14 11 21
0 0 1 -4/5 12/5
0 0 0 1 15.02

Answer 76

Now the coefficient matrix is upper triangular. We're done

Answer 77

\(\large v_5 = 15.02\)

\(\large v_4 - \frac{4}{5}v_5 = 12/5\)

\(\large v_3 - 14 v_4 +11 v_5 = 21\)

\(\large v_2 - 2 v_4 - v_5 = -24\)

Answer 78

simply substitute values,

verify the solution wid wolfram
http://www.wolframalpha.com/input/?i=solve++%7B%7B159%2C-43%2C-126%2C126%7D+%2C+%7B1%2C0%2C-2%2C-1%7D%2C+%7B-1%2C1%2C-12%2C12%7D%2C+%7B0%2C0%2C5%2C-4%7D%7D%7B%7Bw%7D%2C%7Bx%7D%2C%7By%7D%2C%7Bz%7D%7D+%3D+%7B%7B756%7D%2C+%7B-24+%7D%2C+%7B45%7D%2C+%7B12%7D%7D

Answer 79

thanks alott... I got it...! iits lengthy but its the solution :) I will practice more questions like this to make it a good understanding :) thanks @ganeshie8

Answer 80

cool :)

Answer 81

and what about inverses and about crammer rule? I know now I am teasing u ... but if u have time can u please solve this question by crammer's rule and by inverses... just do it not here because i think its difficult to write here... do it on some register and then send me the pic if u are free?

Answer 82

crammars ruleinvolves finding 5 determinants, for a 4x4 matrix

Answer 83

inverse method, we need to find the inverse one time. we can use Gause Jordan method for finding inverse which is very simple

Answer 84

I dont know whats that method.... okey let me not disturb u and I will search gause jordan method and also will search for crammers rule for 4 by 4 or 5 by 5 matrices... just give me the link if easy... :)

Answer 85

If I will be unable to understand the method then I will ask u...okey :)

Answer 86

crammars rule is easy to understand,
but for Gauss Jordan method - we need to do the same row operations we did before

Answer 87

yea sure, just ping me when u may get stuck.. :)

Answer 88

okey I will .... thanks a lot... thats the best response I have got since I joined this site...

Answer 89

thanks too :)