LikeA'13OSS Posted December 14, 2012 Report Share Posted December 14, 2012 Hey can anyone help me with this question? Solve the system of linear equations using the row reduction method: x+2y-2z=53x+2y+z=0-x+y-3z=6Please try to show me the solving as well if you can.thanks Reply Link to post Share on other sites More sharing options...
Nabz Posted December 14, 2012 Report Share Posted December 14, 2012 (edited) Hey can anyone help me with this question? Solve the system of linear equations using the row reduction method:x+2y-2z=53x+2y+z=0-x+y-3z=6Please try to show me the solving as well if you can.thanks Very easy:(1 2 -2) (x) (5)(3 2 1) x (y ) = (0)(-1 1 -3) (z) (6)In order to get the values of x y z, you multiply both sides by the inverse of 3 x 3 matrix on the left side. I'm sure you'll be able to find out on internet how to do so and therefore it would look like:(x) (7 -4 -6) (5)(y ) = (-8 5 7) x (0)(z) (-5 3 4) (6)(x) (7x5)+(-4x0)+(-6x6) (-1)(y ) = (-8x5)+(5x0)+(7x6) = (2)(z) (-5x5)+(3x0)+(4x6) (-1)Therefore, x=-1, y=2, z=-1EDIT: haha i'm sorry for the messed up order, but i hope you get the idea... Edited December 14, 2012 by shad0wboss Reply Link to post Share on other sites More sharing options...
LikeA'13OSS Posted December 14, 2012 Author Report Share Posted December 14, 2012 Hey can anyone help me with this question? Solve the system of linear equations using the row reduction method:x+2y-2z=53x+2y+z=0-x+y-3z=6Please try to show me the solving as well if you can.thanks Very easy:(1 2 -2) (x) (5)(3 2 1) x (y ) = (0)(-1 1 -3) (z) (6)In order to get the values of x y z, you multiply both sides by the inverse of 3 x 3 matrix on the left side. I'm sure you'll be able to find out on internet how to do so and therefore it would look like:(x) (7 -4 -6) (5)(y ) = (-8 5 7) x (0)(z) (-5 3 4) (6)(x) (7x5)+(-4x0)+(-6x6) (-1)(y ) = (-8x5)+(5x0)+(7x6) = (2)(z) (-5x5)+(3x0)+(4x6) (-1)Therefore, x=-1, y=2, z=-1EDIT: haha i'm sorry for the messed up order, but i hope you get the idea...hmm thank you!but isnt this different from the row reduction method we are supposed to use in IB? Because we are supposed to use 3 reduction methods (its called reduced row echelon form) and through this we must reach a matrix that looks like an identity matrix.. any idea on how to use that method? I know the three methods we are allowed to use which are switching the rows of the matrix, multiplying a row with a number and adding rows of the matrix to each other. I am having a hard time using this particular technique, so if youve got any ideas please do share Reply Link to post Share on other sites More sharing options...
CkyBlue Posted December 15, 2012 Report Share Posted December 15, 2012 Damn, I don't think I learned rref until university O.o And finding the inverse of a 3x3 matrices is so time consuming it's awful. You have to compute the determinant, and use the inverse of that multiplied by the adjoint of the matrice. (I don't believe you have to know this in IB2 SL Math) but you do it without knowing because you are expected to know how to compute a 2x2 matrix (determinant, inverse and all that). Unless of course the syllabus changed I sat my math exams in 2011.Row reducing is very similar to how you solve a system of equations. I'll assume you know how it works:1 2 -2 53 2 1 0-1 1 -3 6That is a matrix you have to set up. I will refer to the first row as Eq. 1, the second to Eq. 2, .... Notice the last column is the result of the 3 equations. The key is to obtain a "leading 1" and use that to easily manipulate the matrix to rref.1 2 -2 5 Triple Eq. 1 - Eq. 20 4 -7 15 Eq 1+Eq 30 3 -5 111 2 -2 5 Eq.2 - Eq 30 1 -2 40 3 -5 111 2 -2 5 Eq 3 - Triple Eq 20 1 -2 40 0 1 -11 0 2 -3 Eq1- Double Eq 20 1 -2 40 0 1 -11 0 0 -1 Eq.1- Double Eq 30 1 0 2 Eq. 2 + Double Eq 30 0 1 -1I hope the steps I showed are right, but if they aren't you should be able to see what I'm doing. So the solution column (transposed) is (-1 2 -1) and those values seem to be correct if you plug them back into the original equations. On the left of the solution column is a 3x3 identity matrix.That took me way too long... Reply Link to post Share on other sites More sharing options...
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