n/a
its still not helping me the link you showed me
just cant get it right :(
\[\left[\begin{matrix}a1 & b1 \\ a2 & b2\end{matrix}\right]\left(\begin{matrix}x \\y\end{matrix}\right)=\left(\begin{matrix}c1 \\ c2\end{matrix}\right)\] 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?
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?
where did you get the a1 and b1 from? still dont get it :(((
The equations can be written as y+x=1 y-2x=1 in matrix form \[\left[\begin{matrix}1 & 1 \\ 1 & -2\end{matrix}\right]\left(\begin{matrix}y \\ x\end{matrix}\right)=\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]
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?
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.
@ mani this is what this question asked only to gave two straight lines
@ scsam i did similar to your one you just showed. wasent sure if i was right but seems i was thanks for your help
@bhysen Anytime :)
now is asking for this: Find the inverse of the 2-by-2 matrix A obtained above:
2-by-2 Matrix of A
the inverse of a 2x2 is simple enough if you recall the pattern; AND if its determinant aint zero
ab d -b cd -c a
is that the answer of this The equations can be written as y+x=1 y-2x=1 ?
oh, and divide by determinant
thats the answer in general; you should be able to fill in the specifics from that
|dw:1334678191928:dw| now how should i inverse of the 2-by-2 matrix A?
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