Question

Find a non-zero, two-by-two matrix such that:
[-6 -5] x [__ __] = [0, 0]
[24 20] [__ __] [0, 0]

It's a question that has been asked one year ago by someone else but I can't figure out how they get the matrix :
[-5 -5]
[6 6]

When I try to do it. Here is what I get and where I get stuck :

-6a -5c = 0
-6b-5d= 0
24a-20c = 0
24b-20d = 0

At this point, I just have NO IDEA how to solve these equations. I am sure it's something really easy to do but I still need help.

Answer 1

If you crossed the the matrices correctly and got those equations, solving them simultaneously gets you 0=0?

Answer 2

Well that's the problem, I have no idea what to do. I'm pretty sure it's something simple that I am missing. Here is what I do :

-6a-5c = 0 so a = -(5/6)c
Then I replace the a in -6a-5c=0 with the -(5/6)c. I end up with 0=0..

Sorry for my English.

Answer 3

I see your problem. When you solve one equation for an unknown, you cannot replace it back in the same equation. Try into another one.

Answer 4

I just replaced the a in 24b-20a = 0. I end up with c = (-36/25)b. I still have no idea what I need to do to be honest. Do I have to do this for all the equations and then try to figure out a pattern?

Answer 5

Now you have c. Replace c in other equations. Keep going until you get a value for any unknown. Then start substituting again to get other values.

Answer 6

I'm so sorry but I'm still really confused. I never end with a really value when I do that unless I'm missing something.

a = -(5/6)c
b = -(5/6)d
a = (5/8)c
b = (5/8)d
d = (8/5)b

Answer 7

Oh my. I'm sorry but I can't show you step-by-step to do this now because I'm going to bed. But I think I see a mistake in doing the cross of the matrices. Where did your d go?

Answer 8

Yes I'm sorry I did a mistake on my first post. Here is an explication I found :

"Then working out the top left coefficient of your matrix product will give you a relationship between a and c, so you can eliminate one of them (express a in terms of c, in B), and similarly for the top right. B will now just have two unknowns which can take any value.

You don't need to worry about the second row of A as it is a multiple of the first row."

I am so confused and lost. I remember doing this kind of equations when I was 16 but it has been so long that I feel stupid for not knowing how to simply solve two equations.
Thanks for your time!

Answer 9

No problem

Answer 10

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Answer 11

Ok thank you goformit100. I still didn't get an answer on how to solve so I can't close the question for now sorry.

Answer 12

I will reformulate my question. How to solve these equations :

-6a -5c = 0
-6b-5d= 0
24a-20c = 0
24b-20d = 0

Answer 13

in the first ROW you have of matrix 1 you have
-6 -5
we need -6x-5y=0
two such numbers are x = 5 and y =-6
so we need to make sure if this works for ROW 2 of matrix 1
row two is 24 20
24**5-6*20 = 0
so
5
-6

is the first COLUMN of matrix 2

Answer 14

the same argument holds for row 2