Solving a Matrix Question: -8x-5y-9z=4

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Homework Help Overview

The discussion revolves around solving a matrix equation, specifically the equation -8x-5y-9z=4, and later a system of equations involving variables A, B, and C. The subject area includes linear algebra and matrix manipulation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to express one variable in terms of others due to having a single equation with three variables. They question the correctness of their approach. Other participants discuss the implications of having multiple solutions and suggest checking the validity of the expressions derived.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's attempts. Some guidance has been offered regarding further steps in solving a subsequent system of equations, indicating a productive direction in the conversation.

Contextual Notes

There is a mention of constraints related to the online assignment, where the original poster has only one attempt to submit their answers. Additionally, the transition from one equation to a system of equations introduces complexity that is being navigated in the discussion.

Melawrghk
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Homework Statement



Solve: -8x-5y-9z=4

The Attempt at a Solution



I just need to know if I'm doing this correctly...

There is only one equation given, so I expressed x in terms of y and z and got:
x=-5/8y-9/8z-1/2
then I set y=s and z=t
x=-(5/8)s -(9/8)t -1/2

so I'd enter the answers as:
x = -(5/8)s-(9/8)t-1/2
y = s
z = t

Is that correct? I only have one attempt for the online assignment.

Thanks!
 
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Try checking it yourself. You have three variables and one equation, so there must be multiple solutions as you suggested. You should be able to pick arbitrary values for "s" (y) and "t" (z) to determine your "x". Do they satisfy the given equation?
 
Yeah it looks ok.
 
Thanks! It was right. I have another problem though. This one I'm actually stuck on...

solve the system:
6A-2B+3C=4
1A+6B-6C=20
3A-1B-2C=9

I made the matrix:
6, -2, 3, 4
1, 6, -6, 20
3, -1, -2, 9

And through some simple things I got to:
1, -13, 11, -31
0, 19, -17, 51
0, 0, 5, -14

After which I don't know what to do... Help?
 
Well, you've reduced the matrix to a row-echelon form. Why not perform further row reduction to reduce it to its reduced-row echelon from? Remember that the leading entry of every pivot column of a RREF matrix is 1.
 
Assuming that is correct (I haven't checked the calculations), those now correspond to x- 13y+ 11z= -31, 19y- 17z= 51, and 5z= -14.

Now you can "back substitute". Solve the last equation for z, substitute that into the second equation and solve for y, then substitute those two values for y and z into the first equation and solve for x.
 

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