Question

Can someone take me step-by-step through this problem.

Answer 1

im thinking an augment of the 2 might get you the results

Answer 2

\[\left[\begin{matrix}1 & 1 & 1 &0&1&0\\ 0 & 1 & 1&1&0&0 \\ 0 & 0 & 1&0&0&1\end{matrix}\right]\]

Answer 3

1a + 0b + 0c = 0
1a + 0b + 0c = 1
1a + 0b +0c = 0

a = [0,1,0]
-------------------------

1(0) + 1b + 0c = 1
1(1) + 1b + 0c = 0
1(0) + 1b +0c = 0

b = [1,0,0]
--------------------------

1(0) + 1(1) + 1c = 0
1(1) + 1(0) + 1c = 0
1(0) + 1(0) +1c = 1

c = [-1,-1,1]
-------------------------

F:

0 1 -1
1 0 -1
0 0 1

Answer 4

or the augment yeah

Answer 5

im sure i missed something on my write up

Answer 6

Or is the augment the other way around
\[\left[\begin{matrix}0 & 1 & 0 &1&1&1\\ 1 & 0 & 0&0&1&1 \\ 0 & 0 & 1&0&0&1\end{matrix}\right]\]

Answer 7

F(v) = Ax

or put another way

Av = F(v)

Answer 8

so to find A, it would be some sort of: (v,F(v)) -> ( I ,A )

Answer 9

So the result is this

Answer 10

\[A=\left[\begin{matrix}-1 & 1&0 \\ 1 & 0&-1 \\ 0&0&1\end{matrix}\right]\]

Answer 11

if we run the long version, i see where i goofed

1a = 0
1a = 1
1a = 0

a = [0,1,0]
-------------------------

1(0) + 1b = 1
1(1) + 1b = 0
1(0) + 1b = 0

b = [1,-1,0]
^^
not 0
--------------------------

1(0) + 1( 1) + 1c = 0
1(1) + 1(-1) + 1c = 0
1(0) + 1( 0) +1c = 1

c = [-1,0,1]

Answer 12

your augment is correct, which means that either my augment idea is wrong, or its just not in a suitable order; the long version suits me good tho

{{0, 1, -1,}{1, -1, 0},{0, 0, 1}} * {{1},{0},{0}}
{{0, 1, -1,}{1, -1, 0},{0, 0, 1}} * {{1},{1},{0}}
{{0, 1, -1,}{1, -1, 0},{0, 0, 1}} * {{1},{1},{1}}

Answer 13

maybe the augment has to transpose the columns?

Answer 14

lol, when i transpose v, we get a transpose of what i found

Answer 15

Year.. :)

Answer 16

ah, that does make sense now that i consider it