Linear Algebra - Writing solutions as a span of vectors

J

JaiLit51853

New member
Joined
Apr 20, 2019
Messages
10
Hi there,

I have this 2x5 augmented matrix that represents a homogenous and am supposed to write all solutions as a span of three vectors. (Do not have written question to post).

| 0 1 -1 -1/2 | 0 |
| 0 0 0 0 | 0 |

I know that 0a + b - c - (1/2)d = 0.
B is the leading variable, so a, c, and d are all free variables.

I rewrote the equation into this: b = 0a + c + (1/2)d
I know that a=a, c=c, and d=d.

I then tried to write the solution using vectors:

| a | = | a |
| b | = | c + (1/2)d |
| c | = | c |
| d | = | d |

| 1 | | 0 | | 0 |
= a | 0 | + c | 1 | + d | 1/2 |
| 0 | | 1 | | 0 |
| 0 | | 0 | | 1 |

So therefore the solution in the proper form would be:

{ | 1 | | 0 | | 0 | }
{ | 0 | , | 1 | , | 1/2 | }
{ | 0 | | 1 | | 0 | }
{ | 0 | | 0 | | 1 | }

But my answer is wrong, and I really can't figure out why. I watched lots of videos online to try to figure out where I made mistakes and my method still makes sense to me.
Any comments/help would be greatly appreciated!
 
Last edited:
a, c and d can be any real number but b (NOT B) = c + (1/2)d
So let a=r, c=s and d=t
So the solutions are a=r, b= s + .5t, c= s and d=t
 
JaiLit51853 said:
I have this 2x5 augmented matrix that represents a homogenous and am supposed to write all solutions as a span of three vectors. (Do not have written question to post).
| 0 1 -1 -1/2 | 0 |
| 0 0 0 0 | 0 |
Click to expand...
Your notation is a total mystery to me.
If I were you, then I would post the original question in the complete original wording.
Maybe we can help you, if we can read the question.
 
Jomo said:
a, c and d can be any real number but b (NOT B) = c + (1/2)d
So let a=r, c=s and d=t
So the solutions are a=r, b= s + .5t, c= s and d=t
Click to expand...
Thank you Jomo! I understand this (I probably should've wrote it this way), but am confused as to why the solution set I got when I put in that information is incorrect. Pka, thanks for letting me know. I just realized that the notation could be confusing typed and will rewrite it on paper and post a picture. I will try harder to find the original question and will post it here if I can.
 
Jomo said:
On the 2nd image, the 2nd equal sign is nonsense.
Click to expand...
Oh, why is that? I'm fairly sure it has to go into parametric form after the second equal sign, did I just mess up badly? I did it first and watched this video to check for mistakes but thought it was the same (watch).
 
The middle term is a sum of vectors while the right side is a list of 3 vectors. Why would you think that they are equal? When you add three vectors you get a single vector, NOT a list of three vectors. I know what you meant to write but you did not. Your equal signs MUST be valid!
 
Oh shoot you're right! The equals sign shouldn't be there. I think I forgot to write the last step in the original, the answer I was looking for is just the three vectors. I had to find the form with the scalars attached to the vectors so I could then write the list of vectors.
 
You must log in or register to reply here.
Top