Question

The Matrices

Answer 1

The matruces A, B and C are given by
\[A = \left[\begin{matrix}1 & 5& 6 \\ 2 & -2& 4\\ 1 & -3& 2\end{matrix}\right]\]
\[B = \left[\begin{matrix}-13 & -50 & -33 \\ -1 & -6 & -5\\ 7 & 20 & 15\end{matrix}\right]\]
\[C = \left[\begin{matrix}4 & 7 & -13 \\ 1 & -5 & -1\\ -2 & 1& 11\end{matrix}\right]\]
Determine AB, ABC and hence, deduce \[(AB)^-1\]
Using the result, solve the following system of linear equations

6x+10y+8z=4500
z-2y+z=0
x+2y+3z=1080

Answer 2

@alooy

Answer 3

@mathmale @mathmath333

Answer 4

you should first find the inverse of a*b matrix.

Answer 5

then?

Answer 6

I calculate AB, and then ABC. I got:
\[AB = \left[\begin{matrix}24 &40& 32 \\ 4 & -8& 4\\ 4 & 8& 12\end{matrix}\right]\]
\[and~~ABC = \left[\begin{matrix}72 & 0& 0 \\ 0 & 72& 0\\ 0 & 0&7 2\end{matrix}\right]\]

Answer 7

and \(det(AB)\neq0\) so that AB has inverse. Moreover, \(AB^-AB =I\) It leads me to
\(C = AB^-* \left[\begin{matrix}72 & 0& 0 \\ 0 & 72& 0\\ 0 & 0&7 2\end{matrix}\right]\)
\[or~~AB^-= C* \left[\begin{matrix}72 & 0& 0 \\ 0 & 72& 0\\ 0 & 0&7 2\end{matrix}\right]^-\]
That is for part 1)
for part 2) I don't see any link between the previous part to it. I can solve it independently but not apply the result from previous one. hehehe. I am sorry for my helpless.

Answer 8

Walid: With what do you need help right now? I'd suggest you post just one part of this problem at a time; otherwise postential helpers could get confused regarding where you want them to focus.

Answer 9

@mathmale
The second part is related with the first question. So should follow the flow

Answer 10

What specifically do you need and want help with right now?

Answer 11

Solve linear equations using the first part

Answer 12

@mathmale

Answer 13

@OOOPS Thanks

Answer 14

Point out the specific problem you want to discuss now.

Answer 15

@mathmale

this my problem>> Using the result, solve the following system of linear equations

6x+10y+8z=4500
z-2y+z=0
x+2y+3z=1080

Answer 16

Walid: Have YOU actually found AB and \[(AB)^{-1}\]? Until you have that and understand where it came from, we cannot move forward to the last part of this problem.
Alternatively, we could use matrix row operations to solve the given system of equations and ignore the work we've done before. Your choice. Either you find AB and\[(AB)^{-1}\] or forget about AB and start from scratch to solve the given system using matrices. Which will it be?

Answer 17

yes i have done AB inverse

Answer 18

Plese type out your (AB) inverse here.

Answer 19

72 0 0
0 72 0
0 0 72

Answer 20

But "Ooops" labeled that ABC, not the inverse of AB.

Answer 21

I need to see YOUR calculations of AB and AB inverse, not those of "Ooops".

I can help you solve the given system of equations, but not in the manner the question asks you to follow.

Answer 22

If you have the system

6x+10y+8z=4500
z-2y+z=0
x+2y+3z=1080

you could re-write it in the form of a matrix equation: [A][X] = [B] and to solve this, you'd need to find the inverse of matrix A and then multiply both sides of the matrix equation by the inverse matrix of A. Have you solved any systems of linear equations in this manner before?

Answer 23

yeah i know this method but the problem am facing is the relation between these two parts

Answer 24

I undrstand that you're somehow supposed to use our previous results to solve this system of linear equations, and that your solution must involve finding one or more inverse matrices. Exactly how to make the jump from Part A to solving this given system, I do not know. However, IF you want to attempt to solve the system of linear equations using matrix equations and inverse matrices, I can guide you through that. Or, you could try to find someone else here on OpenStudy who could help you get from Part A to the solution of the given system of linear equations. your choice?

Answer 25

thanks

Answer 26

You're welcome. Have you decided which course of action you're going to take?

Answer 27

look at (AB)/4

Answer 28

then look at your system of equations

Answer 29

the same

Answer 30

then?

Answer 31

set up a matrix equation and solve using inverses

Answer 32

ok

Answer 33

we"ll get the values of x, y and z

Answer 34

@Zarkon

Answer 35

Walid: Could you begin with

6x+10y+8z=4500
z-2y+z=0
x+2y+3z=1080

and identify 'system matrix' [A], variable matrix [X], and constant matrix [B]?

Answer 36

yup

Answer 37

OK. draw or type in [X] and [B]. Then draw or type in [A].

Answer 38

The given system of equations is to be written in the form [A][X]=[B].

Answer 39

yeah this one i know

Answer 40

Good. Do it, please.

Answer 41

To remind you: The solution is found from\[[A]^{-1}[A][X]=[A]^{-1}[B]\]

Answer 42

Very nice! Now find the Inverse of [A].

Answer 43

You may find it easier to write out the inverse of [A] if you put it into the form\[\frac{ 1 }{ 18 }[3x3~integer~matrix]\]

Answer 44

what you have there agrees perfectly with my own result.

If you wish, use your result as is, or, factor out the (1/18).
Multiply your [B] by this \[[A]^{-1}\] Simplify the result, and then you'll be done.

Answer 45

thanks

Answer 46

So...have you been able to solve for [X], which is a 3x1 matrix representing the solution (x,y,z)?

Answer 47

yep

Answer 48

good. Even though we did not solve this problen exactly as we were told to solve it, we do have a solution. If you want, you could compare your most recent work to what you did before and determine whether or not you see any parallels.

Answer 49

I need to get off the 'Net now. Hope this discussion was helpful to you. Best wishes.

Answer 50

Thanks alot
All the best