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it is not C
this takes a bit but not too bad ready?
this is where you say "yeah i am ready even though i hate this even more"
yeah i am ready even though i hate this even more
ok the last line has 1, -2 which means \(z=-2\)
the second to last line has 1, 1, 3 which means \[y+z=3\] we know \(z=-2\) so \[y-2=3\]or \[y=5\]
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 \[x+5a-4=2\]
solve for \(a\) and get your answer you can do it or you want help?
oops i meant "solve for \(x\)"
its B? then
okay 1 or 2 more then i promise ill let you go :'(
i have to say for someone starting out with matrices this is a dumb question but whatever
yeah i always stay up until 3 to help...
I think its D?
@freckles this one is on you, i always screw this up
C is out
i mean matrix C is out
oh maybe not let me shut up
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
so BA is ok right?
okay so can CB not work or
CB ok [2,4][4,1] right?
oh wait so it's E then??
AB will not work since we have a 1 by 3 times a 4 by 1 3 is not 4 BA will work since we have a 4 by 1 times 1 by 3 1 is 1 CB will work 2 by 4 times 4 by 1 because 4 is 4
otherwise this is total donkey work that takes like forever with very little bang for your buck it is for a machine, let the machine do it http://www.wolframalpha.com/input/?i=x-2y-z%3D2%2Cy%2B3z%3D1%2C2x-4y-3z%3D3
you got this?