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Mathematics 17 Online
OpenStudy (anonymous):

Find the solution to the system of equations represented by the matrix shown below. 3 -1 7|53 1 7 1|61 9 1 1|67 A.x = 6, y = 21, z = 54 B.x = 6, y = 7, z = 6 C.x = 42, y = 7, z = 6 D.x = 54, y = 21, z = 54

OpenStudy (anonymous):

Are you just checking answers, or are you not sure how to find it?

OpenStudy (anonymous):

Both.

OpenStudy (anonymous):

Well how can you check your answer and not know how to do it? You'd have no answer to check ;p

OpenStudy (anonymous):

You don't know how to use gauss-jordan elimination to write a matrix in reduced row echelon form to retrieve the solution to a system of linear equations? You can try plugging in values from each multiple choice answer until they fit.

OpenStudy (anonymous):

True, you could find it by process of elimination certainly.

OpenStudy (anonymous):

I was never taught how to do it. Its part of a test im taking..

OpenStudy (anonymous):

That sucks :-( Lets teach you :-D there are video tutorials out there I'll link one to you or I can type the algorithm here.

OpenStudy (anonymous):

Okaay, thank you so much!

OpenStudy (anonymous):

http://www.youtube.com/watch?v=CsTOUbeMPUo

OpenStudy (anonymous):

ughh. this is so confusing :(

OpenStudy (anonymous):

This matrix is a little difficult to do by hand. if you see hard matrices like this in multiple choice you should probably guess it to get choice b (by trying all answers to see if they work). It is faster than elimination To eliminate matrices you follow basic steps to eliminate each column: 1. switch a row containing a leading '1' into the top 2. perform row operations (multiplying row 1 and subtracting row 1 from the values in the other rows) to make all the other elements in the first column 0 3. switch columns and do the same steps but for the second row to have a leading 1 for this particular matrix you can follow these steps: Eliminating the first column 1. switch rows 2 and 1 (so the top row reads 1 7 1 61) 2. multiply row 1 by 3 and subtract row 2 by row 1 (so row 2 reads 0 -22 4 -98) 3. multiply row 1 by 9 and subtract row 3 by row 1 (so row 3 reads 0 -62 -8 -410) now you have a matrix that reads [ 1 7 1 | 61 ] [ 0 -22 4 | -98 ] [ 0 -62 -8| -410] next step is to do something like divide row 2 by -22, and then subtract it from row 3 (after multiplying by -62). you should finally get something like 1 0 0 6 0 1 0 7 0 0 1 6 in the end

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