Question

Relearning a little of Linear Algebra...

Question in a bit...

Answer 1

oki

Answer 2

Let \(A=\left[\begin{matrix}1 & 2 \\ 3 & 4\end{matrix}\right],\ B=\left[\begin{matrix}1 & 3 \\ 4 & 5\end{matrix}\right]\ and \ C=\left[\begin{matrix}1 & 4 \\ 6 & 8\end{matrix}\right]\).
Is the set \(\theta=\{A,B,C\} \) linear independent?

The procedure is the same as done for vectors in \(R^{n}\) or is there a catch?

Answer 3

I mean, I'd go for \(\alpha A+\beta B+\gamma C=\vec{0}\) and then solve a system consisting of a b and c.

Answer 4

Yes, thsts what I think, but I am not 100% sure

Answer 5

oops, a system consisting of \(\alpha,\beta\ , \gamma\)

Answer 6

Thats the only way I learned so I guess it works fro every situation

Answer 7

The row echelon form for the augmented matrix I found is given by:
\[\left[\begin{matrix}1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 2 & 0\end{matrix}\right]\]

Which only yields the trivial solution... Did I get it right?

Answer 8

Noone? :(

Answer 9

I guess I got it right. Closing the question in 10 minutes if noone responds.

(Wolfram answer: https://www.wolframalpha.com/input/?i=linear+dependence+ {1,2,3,4},{1,3,4,5},{1,4,6,8})

Answer 10

Yeah you're right the matrix implies what you stated

Answer 11

Not that
let a[A] +b[B]+c[C]=0
we need find a, b,c such that
a+b +c =0 (the first entry T_11)
2a =3b+4c=0(the second entry T_12)
3a+4b+6c=0
4a+5b+8c=0
solve for a,b,c

Answer 12

Yes, a=b=c=0

Answer 13

Hence they are linearly independent.

Answer 14

I managed to find the linear system. I just laid out the row echelon form of the augmented matrix

Answer 15

that is a[ A]+b[B]+c[C]=0 iff a=b=c =0

Answer 16

I don't solve it, I use calculator to find rref

Answer 17

Oh well, guess I got it right, then!

Answer 18

Thanks for checking!