Mathematics
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OpenStudy (anonymous):
Matrix Multiplication
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OpenStudy (anonymous):
OpenStudy (anonymous):
@KendrickLamar2014
OpenStudy (kendricklamar2014):
Can you try to solve it or do you need me to show you?
OpenStudy (anonymous):
give me a sec to look at it
OpenStudy (kendricklamar2014):
Ok
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OpenStudy (anonymous):
Ok so would I multiply each number in the first matrix by 2?
OpenStudy (kendricklamar2014):
all the numbers in the matrix are the variable x. So you would do: 2 x(# in the matrix) + 2
OpenStudy (anonymous):
ohhhhhh! ok give me a sec
OpenStudy (kendricklamar2014):
:)
OpenStudy (anonymous):
so after multiplying each number by 2, would I add 2 to each of them?
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OpenStudy (anonymous):
how does the + 2 fit in?
OpenStudy (kendricklamar2014):
Let me do the 1st one:
The 1st number in the Matrix is 2.
So you would do:
2*2+2
4+2
6
OpenStudy (anonymous):
sorry, like this?
6 -14
-6 6
OpenStudy (kendricklamar2014):
Yes
OpenStudy (anonymous):
ok but there's an = sign between them...
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OpenStudy (anonymous):
so what answer am I looking for?
OpenStudy (isaiah.feynman):
How about writing the matrices as just A and B, and solving for x as in algebra.
OpenStudy (kendricklamar2014):
^
OpenStudy (anonymous):
whoa ok I'm lost already
OpenStudy (isaiah.feynman):
\[2X+2A = B\]
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OpenStudy (kendricklamar2014):
ohh i was doing it wrong :( sorry
OpenStudy (anonymous):
its fine @KendrickLamar2014 you were trying :)
OpenStudy (isaiah.feynman):
Then we can solve for x from here, like so...\[X = \frac{ 1 }{ 2 }\left( B-2A \right)\]
OpenStudy (anonymous):
ok... let me try to work it out...
OpenStudy (isaiah.feynman):
We know A and B. So, do necessary substitution.
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OpenStudy (anonymous):
so the answer is B??
OpenStudy (anonymous):
0 -11
-3 -6
OpenStudy (isaiah.feynman):
I didn't solve it actually.
OpenStudy (anonymous):
oh haha well I think that's the right answer. thank you
OpenStudy (isaiah.feynman):
Welcome!