❖ Gaussian Elimination ❖

okay here I'm going to use Gaussian elimination to solve a system of equations we've got X plus y minus Z equals 9 y plus 3z equals 3 negative x minus 2z equals 2 so the first thing I'm going to do is write down my Augmented matrix so from the first row our coefficients are positive 1 positive 1 negative 1 our constant is 9 our second row we have 0 X we have 1 Y we have 3 Z and our constant is 3 and from our last row we have negative 1 X we have 0 Y negative 2z + our constant is positive 2 so what we're gonna do is our goal in this case is we want to write this in row echelon form so the idea is we're gonna write the matrix so that it's in triangular form and where the leading coefficient in each row is equal to 1 okay so to do this I look at my first row I want the first nonzero entry in the first row to be a 1 so in this case we've got a positive 1 here and again the idea is if you kind of look along the diagonal forget about the constants if you look along the diagonal XYZ we're gonna make the entries below that into zeros we're gonna make the the constants all equal to 1 that's that's what we're shooting for here okay once we do that we're gonna use some back substitution so again my goal eventually is to get zeros here and I want ones along the diagonal that's that's our that's what we're going for okay so that'll kind of be step one here okay so to get 0 and the in the third row first column I'm just going to take Row one and add that to Row 3 and that's going to replace my Row three I like to use the little arrow to indicate which row is changing there's tons of arithmetic in these problems so anything I can you know I can do to help keep myself organized I always found that helped so I'm not changing the first row not changing the second row either and now I'm going to do my arithmetic so 1 plus negative 1 will be 0 that's what we wanted 1 plus 0 will be 1 negative 1 plus negative 2 will be negative 3 and then 9 plus 2 will be positive 11 ok so again we want ones along the diagonal again forgetting about the the column with the constants well we do have a 1 in our second row second column so that's good we said we wanted to get zeros everywhere below that diagonal ok everywhere below that diagonal so I need to change the entry in the third row second column I want to make that into a zero as well well if we just add Row two and Row three that's not going to do the trick but if we multiply the second row by negative one and then add that to our third row I think that'll do it for us so let's see here so again I'm not changing the first row at all I'm not changing the second row either now I'm gonna change the third row again by doing the arithmetic so negative 1 times 0 plus 0 that's just 0 negative 1 times 1 will be negative 1 if we add negative 1 to positive 1 we'll get 0 which is exactly what we wanted negative 1 times 3 is negative 3 negative 3 plus negative 3 will give us negative 6 negative 1 times 3 again will be well negative 3 negative 3 plus 11 will be positive 8 and now we've got our third row all right here getting close the last thing again we want to do is we want to make our our first nonzero entry in the third row which is our negative 6 we want to make that into a positive 1 well what I'm gonna do to the third row is simply you can think you know I would divide by negative 6 we don't really think about division so much we always talk about multiplying by nonzero constants so I'm gonna multiply the third row by negative 1 sixth and that's gonna give me my new third row so in essence you're just dividing the third row by negative 6 again not doing anything to the first row not doing anything to the second row negative 6 divided by negative 6 would be positive 1 8 divided by negative 6 well that would be 8 over negative 6 and we can reduce 8 over 6 into 4/3 again not forgetting the negative ok so now we've got our matrix in the form that we want and what we do now is we go back and write the corresponding set of linear equations and then we start using substitution back substitution to solve so recall this is the column corresponding to X the column corresponding to Y the column corresponding to Z and then our constants so we've really got the system of equations 1x plus 1 y -1 Z equals 9 from our second row we've got 0 X plus 1 y plus 3 Z equals 3 and from our our last row we simply have that positive 1 z equals negative 4/3 ok so now we know what ze rules the idea is we can take that and substitute that into the second equation solve for y then we'll have both y&z then we can substitute that into the first equation and solve for X ok so using our second equation we have y plus 3 times Z that's negative 4/3 that equals positive 3 well the 3s would simply cancel out so we would have Y minus 4 equals 3 and if we add 4 to both sides we'll get that y equals positive 7 ok so now I'm going to take these two values and simply put those back into our first equation so again we had X plus y which is 7 minus the value of Z which is negative 4/3 and that's gonna equal positive 9 all right so very close here we've got X plus 7 we would have 2 plus 4/3 equals 9 and you could always get common denominators and add subtract fractions I'm gonna get rid of the fractions or it's a fraction by multiplying both sides by positive 3 so we'll have to distribute that we'll get three x three x seven as positive 21 three x four thirds we'll just be positive four nine times three will be 27 let's see we would have 3x plus 21 plus four will be 25 and if we subtract 25 from both sides we'll get that 3x equals two and if we divide both sides by three we'll get that x equals two-thirds so now we've got our solution it says the solution to the system of equations would be when x equals two-thirds we said that y equals 7 and Z equals negative 4/3

Channel: patrickJMT

Views: 2,535,944

Rating: 4.8662729 out of 5

Keywords: gaussian, elimination, triangular, upper, echelon, back, substitution, math, linear, algebra, middle school, high school, online class, online university, online college, tutorial, free, textbook, Gaussian Elimination, Linear Algebra (Field Of Study)

Id: 2j5Ic2V7wq4

Length: 8min 55sec (535 seconds)

Published: Thu Jul 12 2012