Question

Please, check my stuff:
determine the solution set to the system Ax =0 for the given matrix A

Answer 1

\[\left[\begin{matrix}1-i&2i\\1+i&-2\end{matrix}\right]\]
Multiply row 1 by -i , I get
\[\left[\begin{matrix}-1-i&2\\1+i&-2\end{matrix}\right]\]
add row1+row2.

Answer 2

so, (1+i )x= 2y
the solution set will be S= { (1+i)/2 t, t| t in R}
but it's wrong. I don't know why??

Answer 3

\[\left[\begin{matrix}1-i & 2i \\ 1+i & 2\end{matrix}\right]\left(\begin{matrix}a \\ b\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]

multiply out and solve for a & b

Answer 4

(1-i)(-i)=-i +i^2 = -1-i

Answer 5

check if det[A] = 0 because if it's not 0 then the 0 vector is only solution.

Answer 6

so, what 's wrong with mine?

Answer 7

nope, it's not that, friend.

Answer 8

? it's not what?

Answer 9

yeah, det[A] = 0 so there's a non-trivial solution.

Answer 10

since the matrix has one 0 row, therefore, there is one free variable, the solution set is infinite set . the result from book said that it is S ={ (1-i)t,t|t is in R}, not as what I did

Answer 11

take a row...
(1+i)x = 2y solve for y.

y = (1+i)/2 * x

Answer 12

let x = t, then y = [(1+i)/2]t

Answer 13

the same with mine, but it's not right. so. that's why I want you check what's wrong.

Answer 14

it should be the same thing... to check, just solve for x in terms of y...

y=(1+i)/2 * x => (1+i)x=2y => x = 2y/(1+i) now to divide mult by conjugate...

\[ x=\frac{2y\cdot(1-i)}{1+1}=(1-i)y\]

now let t = y then x = (1-i)t

Answer 15

make sense?

Answer 16

give me time to read, does my requirement make sense?

Answer 17

yur requirement? not following

Answer 18

just joke. you just post and then asked me "make sense?" so I required time to read. hihihi

Answer 19

yeah, sorry... please take your time and let me know.

Answer 20

ok, it makes sense to me now. thank you.

Answer 21

you're welcome... i forget that i've already worked it out and of course it makes sense to me and then I forget you've not had the time to look and digest.

sorry!

Answer 22

hehehe... that's my fault. thanks again

Answer 23

no... it's mine!