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OpenStudy (anonymous):
OpenStudy (anonymous):
ihatethisevenmoremore
OpenStudy (anonymous):
it is not C
OpenStudy (anonymous):
this takes a bit but not too bad
ready?
OpenStudy (anonymous):
this is where you say "yeah i am ready even though i hate this even more"
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OpenStudy (anonymous):
yeah i am ready even though i hate this even more
OpenStudy (anonymous):
ok the last line has 1, -2 which means \(z=-2\)
OpenStudy (anonymous):
the second to last line has 1, 1, 3 which means \[y+z=3\] we know \(z=-2\) so
\[y-2=3\]or
\[y=5\]
OpenStudy (anonymous):
the first line has 1, a, 2, 2 which means
\[x+ay+2z=2\] we know \(y=5\) and \(z=-2\) so we have
\[x+4a+2\times (-2)=2\] or
\[x+5a-4=2\]
OpenStudy (anonymous):
solve for \(a\) and get your answer
you can do it or you want help?
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OpenStudy (anonymous):
oops i meant "solve for \(x\)"
OpenStudy (anonymous):
its B? then
OpenStudy (anonymous):
yeah B
OpenStudy (anonymous):
okay 1 or 2 more then i promise ill let you go :'(
OpenStudy (anonymous):
i have to say for someone starting out with matrices this is a dumb question but whatever
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OpenStudy (anonymous):
yeah i always stay up until 3 to help...
OpenStudy (anonymous):
OpenStudy (anonymous):
I think its D?
OpenStudy (anonymous):
@freckles this one is on you, i always screw this up
OpenStudy (anonymous):
C is out
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OpenStudy (anonymous):
i mean matrix C is out
OpenStudy (anonymous):
oh maybe not let me shut up
OpenStudy (freckles):
if we do a m by n times a n by k
the we are cool because this will result in a m by k matrix
you just have to make sure
the inner numbers match if you know what I mean