Solve by Echelon Method a. 4x - y + 3z = -2 3x + 5y -z = 15 -2x + y + 4z = 14 b.

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a. We note: (a1) 4x - 1y + 3z = -2 (a2) 3x + 5y - 1z = 15 (a3) -2x +1y +4z = 14 I note multiplication by a star * The echelon or row echelon is a technique where you reduce the equations as follows: Take the two top equations: (a1) 4x - 1y + 3z = -2 (a2) 3x + 5y - 1z = 15 If I multiply the second equation by 4/3 (everywhere): (a2) (3*4/3)x +(5*4/3 )y -(4/3) z = (15*4/3) or (a2) 4x +20y/3 - 4z/3 = 60/3 Now if we do (a2) minus (a1) the 4x are eliminated: (a2-a1) 4x - 4x +20y/3+1y - 4z/3 -3z = 60/3 + 2 which simplifies as y is 3y/3 and 3z is 9z/3): (a2-a1) 23y/3 - 13z/3 = 22 and multiplying this by +3 on both sides it becomes: (a2-a1) 23y - 13z = 66 Now we do the same thing with (a1) and (a3) again to eliminate x, multiplying (a3) by 2, it becomes: (a1) 4x - 1y + 3z = -2 (a3) -4x +2y +8z = 28 Then adding (a1) to (a3): (a1+a3) 4x - 4x -y + 2y + 3z + 8z = -2 + 28 (a1+a3) y + 11z = 26 Now we have only 2 equations and 2 variables: (a2-a1) 23y - 13z = 66 (a1+a3) y + 11z = 26 and we do that again, multiplying the bottom equation by 23: (a2-a1) 23y - 13z = 66 (a1+a3) 23y + 253z = 598 Subtracting one equation from the other gives: 266z = 532 Or z = 532/266 => z = 2 If we carry this value into (a1+a3) where y + 11z = 26 we get: y + 22 = 26 or y = 4 Finally we carry these values for y and for z into (a1) where 4x - 1y + 3z = -2 we get: 4x – 4 +6 = -2 or 4x = -4 which means x = -1 Verify by putting these values into the 3 equations and you should see they are all balanced. b. Same thing but it seems we do not have enough equations. We have on the website: (b1) x + 2y + 3z = 11, and (b2) 2x – y + z = 2 Normally with 3 variables, you can only solve it if you have (at least) 3 equations. But we can start: Multiply (b1) by 2 gives: (b1) 2x + 4y + 6z = 22 And doing (b1) – (b2) gives: (b1-b2) 2x – 2x + 4y –(-y) + 6z - z = 22 - 2 (b1-b2) 5y + 5z = 20 If you have a third equation, you can finish this yourself … If not the system is indeterminated, with z = (20-5y)/5 or z = 4-y, which can be carried in both equations and make it a 2 equations 2 variables system, but not solvable (the 2 equations are equivalent or dependant on each other, meaning the second one brings no new information): (b1)’ x + 2y + 3(4-y) = 11 simplifies to (b1)’ x – y = -1 (b2)’ 2x – y + (4-y) = 2 simplifies to (b2)’ 2x – 2y = -2 which is the same as (b1)’ times 2. Good luck.

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