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Mathematics 23 Online
OpenStudy (anonymous):

practice for my test can someone please check if i got the answers right please

jimthompson5910 (jim_thompson5910):

I'm checking them one by one and so far #1 and #2 are correct. Nice work

OpenStudy (anonymous):

thanks i think the last one wrong

jimthompson5910 (jim_thompson5910):

#3 is correct

jimthompson5910 (jim_thompson5910):

and #4 is correct

jimthompson5910 (jim_thompson5910):

#5 is incorrect

jimthompson5910 (jim_thompson5910):

#6 is correct

jimthompson5910 (jim_thompson5910):

#7 is correct

jimthompson5910 (jim_thompson5910):

and #8 is incorrect

OpenStudy (anonymous):

can you help me with 5 and 8 explain them to me please?

jimthompson5910 (jim_thompson5910):

Ok we'll start with #5 Let's say we had one equation ax+by = c now let's say that we multiply both sides of this equation by some number k to get ax+by = c k(ax+by) = k*c kax + kby = kc Would you agree that ax+by = c and kax + kby = kc have the same solutions (ie they intersect each other at an infinite number of points)?

OpenStudy (anonymous):

yeah

jimthompson5910 (jim_thompson5910):

So if we can find the determinants D, Dx and Dy, we can determine what they equal when we get an infinite number of solutions (which is what #5 is asking for)

jimthompson5910 (jim_thompson5910):

So we have the system ax+by = c kax + kby = kc which means that to set up the coefficient determinant, we write the coefficients in the matrix like so | a b | | ka kb | What is that determinant?

jimthompson5910 (jim_thompson5910):

Do you know how to find the determinant above?

OpenStudy (anonymous):

no

jimthompson5910 (jim_thompson5910):

Simply multiply the stuff in the main diagonal (a and kb) and subtract off the terms in the off diagonal (ka and b) So a*kb - ka*b = akb - akb = 0 This means D = 0

jimthompson5910 (jim_thompson5910):

You found this, but this isn't the whole story.

OpenStudy (anonymous):

soo....D, Dx and Dy will equal zero.

jimthompson5910 (jim_thompson5910):

Yes, here's why to find Dx, replace the first column with the column c and kc to get | c b | | kc kb | then compute the determinant Dx = c*kb - kc*b = ckb - ckb = 0 So Dx = 0

OpenStudy (anonymous):

ohh i get it :) thankss

jimthompson5910 (jim_thompson5910):

same thing for Dy, but now the second column | a c | | ka kc | Dy = a*kc - ka*c = akc - akc = 0 So Dy = 0 as well

jimthompson5910 (jim_thompson5910):

that's great

jimthompson5910 (jim_thompson5910):

now onto #8

jimthompson5910 (jim_thompson5910):

write the coefficients in a matrix | 4 -3 | | 2 1 | Now replace the second column with the right hand side to get | 4 -14 | | 2 -2 | and then compute the determinant Dy = 4*(-2) - 2*(-14) Dy = -8 + 28 Dy = 20

jimthompson5910 (jim_thompson5910):

So somewhere along the way, you cut something in half...that's my guess anyway.

OpenStudy (anonymous):

ohh i understand that I did wrong

jimthompson5910 (jim_thompson5910):

that's great

OpenStudy (anonymous):

thats for your help @jim_thompson5910

jimthompson5910 (jim_thompson5910):

you're very welcome

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