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Mathematics 13 Online
OpenStudy (anonymous):

n/a

OpenStudy (anonymous):

its still not helping me the link you showed me

OpenStudy (anonymous):

just cant get it right :(

OpenStudy (mani_jha):

\[\left[\begin{matrix}a1 & b1 \\ a2 & b2\end{matrix}\right]\left(\begin{matrix}x \\y\end{matrix}\right)=\left(\begin{matrix}c1 \\ c2\end{matrix}\right)\] a1, b1 are the coefficients of x,y in the first and a2,b2 in the second equation. c1 is the constant term in the first and c2 in the second. ok?

OpenStudy (mani_jha):

sorry, a1 is the coefficient of y and b1 is the coefficient of x. same for a2 and b2 in the second equation. Got_it?

OpenStudy (anonymous):

where did you get the a1 and b1 from? still dont get it :(((

OpenStudy (anonymous):

The equations can be written as y+x=1 y-2x=1 in matrix form \[\left[\begin{matrix}1 & 1 \\ 1 & -2\end{matrix}\right]\left(\begin{matrix}y \\ x\end{matrix}\right)=\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]

OpenStudy (mani_jha):

It is a general form. If a1x+b1y=c1 and a2x+b2y=c2 are two linear equations, a1,b1 are the coefficients of x and y in the first equation and a2,b2 in the second.. The corresponding equations: x+y=1 y-2x=1 What are the values of a1,b1 and a2,b2 now?

OpenStudy (anonymous):

write the co-eff of y in 1 column of the matrix and the co-eff of x in the very next column of the same matrix. so now u have the first matrix. The next matrix will hav e 1 column of y and x. these two matrices after multiplication will another matrix of 1 column having the constants as depicted by R.H.S of the first two equations.

OpenStudy (anonymous):

@ mani this is what this question asked only to gave two straight lines

OpenStudy (anonymous):

@ scsam i did similar to your one you just showed. wasent sure if i was right but seems i was thanks for your help

OpenStudy (anonymous):

@bhysen Anytime :)

OpenStudy (anonymous):

now is asking for this: Find the inverse of the 2-by-2 matrix A obtained above:

OpenStudy (anonymous):

2-by-2 Matrix of A

OpenStudy (amistre64):

the inverse of a 2x2 is simple enough if you recall the pattern; AND if its determinant aint zero

OpenStudy (amistre64):

ab d -b cd -c a

OpenStudy (anonymous):

is that the answer of this The equations can be written as y+x=1 y-2x=1 ?

OpenStudy (amistre64):

oh, and divide by determinant

OpenStudy (amistre64):

thats the answer in general; you should be able to fill in the specifics from that

OpenStudy (anonymous):

|dw:1334678191928:dw| now how should i inverse of the 2-by-2 matrix A?

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