what is the solution of the matrix ?

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what is the solution of the matrix ?

Mathematics
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do you have to do it with matrices?
um i think soo , would you do this the same way you would do it if it was addtion ?if that makes sense

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addition *
ill show you what im talking about
this is what i am talking about would i do it the same but just multiply ?
1 Attachment
not really getting your point inverse of 2x2 matrix can be computed like this |dw:1433291928663:dw|
i see your point, you need inverse and them multiplciation operation of matrices there is not addition here
$$\Large {A}\mathbf{ x} = \mathbf{b} \\ \Large \mathbf{ x} = A^{-1} \mathbf{b}$$
I know ...... :(((((
try to find the determinant first
now how would i find that , i usually use a online matrix calculator to find the determinant so i wouldn't know how to find on my own
ok the determinant is computed like this |dw:1433292400098:dw|
ok
you multiply ad and bc and then subtract
ok thats all ?
do the same with your matrix
no not finished that's only how to do determinant
ok im writing this down
ok
cross multiply ?
yeah seems like cross multiply then take the difference
so for me it would me 9*1 &4*2 for A and for X it would be -9*-8 and -6 *-1 right?
or did you mean multiply A with X ???
so here \[A=\begin{bmatrix}9&4\\ 2&1\end{bmatrix}\] \[\det(A)=9(1)-2(4)=9-8=1\]
oh no what are you taling about? are you following what i'm saying haha
yeah
9*1 i did that 2*4 i did that too
we are looking for X and i used what perl did X=Ainverse times B
i know what i did wrong i forgot to subtract ..
it was AX=B then we multiplied A^-1 and we got X=A^{-1}B now we are in the process of computing A inverse that is why we need determinant of A
no seems you missed the whole poiint from what i understand from your last reply
no .. i didn't you said cross multiply then subtract right ???
that's only for determinant
i did that for you already see my replay above
\(\color{blue}{\text{Originally Posted by}}\) @xapproachesinfinity so here \[A=\begin{bmatrix}9&4\\ 2&1\end{bmatrix}\] \[\det(A)=9(1)-2(4)=9-8=1\] \(\color{blue}{\text{End of Quote}}\)
ok so what do have to do next then ?
find x ?
|dw:1433293368759:dw|
yup
so we do 1/det A so 1/1 =1 here we didn't need to worrt about it since it is just 1
ok
next step is to do a switch|dw:1433293517899:dw|
you take A and switch a and d b and no switch you just change the sign
ok i get that but where do the negatives come from
it is a rule you need to read about it on your notes
what kind of class are taking
algebra 2
hmm really? i don't really know how much you know about matrices
hmm this really needs some time to get accustomed to! it may be fairly easy, but needs a lot of practice you might want to look at some video tutorial on youtube
okay , i will
once you feel comfortable then tackle problem it is no helping to do problem you are not ready to do be real, no matter how much i explain to you here, you will need the lecture first here open study will not give you the lecture rather help assist you with difficulties you are having after knowing the lecture and know fairly enough on the subject
i will do your problem now, but next problem you need to try your hardest :)
okay
let me do it on paper ans take a picture
ok
almost hehe
kay
before the last line i preformed matrix multiplication
is it readable ?
yeah it is . -5 ,26 -9 and -60 correct right
there is no -9
oh ,
that's just brackets
that should match C
oh , my bad it looked like a negative for a second but yes it is C
okay
hope that helps
it does ,,,,, THANKS ..
my pleasure

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