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Mathematics 19 Online
OpenStudy (anonymous):

Which product represents the solution to the system?

OpenStudy (anonymous):

OpenStudy (anonymous):

First do this :(this is not a system !) Look : How can u answer from it : 8.x=16 x=?

OpenStudy (anonymous):

Now we should be have : (3,2)/(11,24)=? @ejune420:Understand ?!:)

OpenStudy (anonymous):

@ejune420:Is my answer fall?

OpenStudy (anonymous):

all the answer choices are matrices

OpenStudy (anonymous):

OK ! I understand it fall :( Excuse me :(

OpenStudy (anonymous):

Did u get it the answer?

OpenStudy (anonymous):

What the problem?

OpenStudy (jdoe0001):

@ejune420 do you know how to get the inverse matrix for the 2x2 matrix there?

OpenStudy (anonymous):

@jdoe0001 yeah I know how to get the inverse

OpenStudy (jdoe0001):

ok well, then the "solution" for it will be \(\large \begin{array}{cccl} A& X & B\\ \begin{bmatrix} 1& 1\\ 2& 4 \end{bmatrix}& \begin{bmatrix} x \\ y\end{bmatrix}=& \begin{bmatrix}3\\2\end{bmatrix}\\ \textit{solution will be at }\\ A^{-1}\times B \end{array}\)

OpenStudy (jdoe0001):

the result will be a 1x2, just like the "X" matrix and will equate the variables

OpenStudy (anonymous):

thanks a bunch for the help, I'm gonna try to work it out now :)

OpenStudy (jdoe0001):

yw

OpenStudy (jdoe0001):

so the discriminant of the 2x2 "A" matrix is 2, so the inverse will be \(\large A= \begin{bmatrix} 1& 1\\ 2& 4 \end{bmatrix} A^{-1} = \cfrac{1}{2}\begin{bmatrix} 4& -1\\ -2& 1 \end{bmatrix}\)

OpenStudy (anonymous):

I got it :)

OpenStudy (jdoe0001):

determinant rather... is 2

OpenStudy (jdoe0001):

ok :)

OpenStudy (anonymous):

thanks so much!!:)

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