find a basis for the null space of A (see attachment)
\[\left[\begin{matrix}1 & -1 &3 \\ 5 & -4 & -4\\7 &-6 &2\end{matrix}\right]\]
@camoJAX this time, it's me. help, please
trying to see now
taking longer than i thought, still going though
First find it's row reduced echelon form.
yep and you'll get \[1x _{1}-1x _{2}+3x _{3}=0\] \[1x _{2}-19x _{3}=0\] still working though
yes, I got it as \[\left[\begin{matrix}1 & -1 &3 \\ 0 & 1 & -19\\0 & 0 & 0\end{matrix}\right]\]
bottom row will be all zeros
@camoJAX bingo.
how to make matrices on here?
stack -22,19, 1 on each other and thats the answer
go to equation right below this box. you can see many symbol, choose matrix symbol, you will have something like ? ? .. whenever see? put the number there and & ? put number at ? if you need 3 terms add 1 & symbol and then type the number there. if you want to take 1 more row, \\ and repeat
To make a matrix type: \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} You use & to delimit columns and \\ to delimit rows.
@camoJAX 16 not 22
\[\left(\begin{matrix}16 \\ 19\\1\end{matrix}\right)\]
my question starts from now on. I got it but is it the null space or we have to break it down to 3 vectors as
You have one free variable, so your null space is just one column
yep dont know why i was thinking negative plus a negative and made the 19 positive then subtracted it from -3. my bad
\[\left(\begin{matrix}16 \\ 0\\0\end{matrix}\right)\left(\begin{matrix}0 \\ 19\\0\end{matrix}\right)\left(\begin{matrix}0 \\ 0\\0\end{matrix}\right)\]
\[ x_1 = x_2 - 3x_3 = 19x_3-3x_3=16x_3 \\ x_2 = 19x_3 \]
meant neg x neg lol
thanks a lot both friends. so, since I have 1 free variable, 1 column basis in null space. if I have 2 , 2 columns basis. is it correct?
Yes, one column per free variable.
agree
thank you very much. "The power of cooperation"
@camoJAX which class do you have now?
i'm still on Cal homework right now
cal3? which school?
yea Cal 3, Lamar Univ..kinda little,not too big school in TX
still better than me, I am in community college.
I am struggling with 3 math courses this semester. I have cal3, linear algebra and discrete
that doesn't mean anything just depends what you do with your knowledge
i'm taking those too except discrete
my discrete is level 2, too many things to do. terrible
i never heard of discrete but just looked it up and I pity you bro (...or sis) lol
sister. no choice for me. my major is math. how can I avoid math courses?
what is your major?
true, sometimes math can be difficult but I love it when I'm able to solve a problem.
Civil Engineering
Lol. not lesser than me in math. for sure
ok, being fan each other to get help whenever we need, deal?
definite deal
hey , my pro gives me a challenge question, deal after spring break, you can take a look and if you have time, think about helping me. i didn't have the solution yet. the problem is : why the procedure to get transition matrix from B to B' work. show your proof.
it means P from B-->B' is |B'|B|----> | I | P from B -->B'|. why it works.
the question is due after the break? your break is March11-15?
no need to press yourself. It is just extra credit, i don't need it that much. my break from 3 to 10. I am during break time. if not, how can i have time to play here
ohh you're on break now. I thought you probably just got on this site whenever. thats what i do
i wanna know the answer now, i'll be thinking about it though
too bad. I asked, pointed at particular people who has smartscore at top 99 -100 all of them say " sorry, I cannot help" or "sorry,it 's not my strong".
i can ask my prof if I never figure it out
ok. it's too late, I have to go to bed now. nice to meet you. whenever you got it, send me message.
cool, night. nice to meet you also
bye bye
Join our real-time social learning platform and learn together with your friends!