Find the linear combination of the vectors
question #50
augment and rref
ya but it is becoming an ugly anser
|its spose to, these things are never pretty :)
can you type up the first one for me, i have trouble trying to switch back and forth and keep things in order
A:= u1 u2 u3 Ax = v x = A-1 v = | A | v|
what do u mean?
give me the u1 u2 u3 and v so that i can work with them; or else im sure to botch it up between switching back and forth from here to there
ok will do
u1= (1,3,2,1) u2=(2,-2,-5,4) u3= (2,-1,3,6) v= (2,5,-4,0)
\[\begin{pmatrix}u_1&u_2&u_3\\1&2&2\\3&-2&-1\\2&-5&3\\1&4&6\\\end{pmatrix}\begin{pmatrix}x\\x_1\\x_2\\x_3\\x_4\end{pmatrix}=\begin{pmatrix}v\\2\\5\\-4\\0\end{pmatrix}\] \[\begin{pmatrix}x\\x_1\\x_2\\x_3\\x_4\end{pmatrix}=RREF\begin{pmatrix}u_1&u_2&u_3&|&v\\1&2&2&|&2\\3&-2&-1&|&5\\2&-5&3&|&-4\\1&4&6&|&0\\\end{pmatrix}\]
thats what I mean; but wolfram make light work of it :)
ummm y did you add the the u1 u2 and u3 ontop?
Is that part of the matrix
just to clairify what was what; its not part of the actual process tho, just references
ohhhh ok i see
Thanks amistre :D
LOL u were born in the same yr as my mom LOL
and my x vector is one component to many :)
what year was I born :)
1964
nope .... thats just an old aol thing thats stuck with me thru the years :)
hehe
so, according to the rref on the wolf: x = <2,1,-1>
yup i see thanks amistre
youre welcome; and i was born in 1972 ;) or so ive been led to believe
hehe so you are not much younger than my mom :D
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