Sunday, October 13, 2013

`x - 4y + 3z - 2w = 9, 3x - 2y + z - 4w = -13, ` `-4x + 3y - 2z + w = -4, -2x + y - 4z + 3w = -10` Use matricies to solve the system of...

`x-4y+3z-2w=9`


`3x-2y+z-4w=-13`


`-4x+3y-2z+w=-4`


`-2x+y-4z+3w=-10`


The above system of equations can be represented by the coefficient matrix A and right hand side matrix b as follows:


A=`[[1,-4,3,-2],[3,-2,1,-4],[-4,3,-2,1],[-2,1,-4,3]]`


b=`[[9],[-13],[-4],[-10]]`


The augmented matrix can be written as,


`[[A,b]]=[[1,-4,3,-2,9],[3,-2,1,-4,-13],[-4,3,-2,1,-4],[-2,1,-4,3,-10]]`


Now lets, perform the various row operations to bring the above matrix in the row-echelon form,


Rewrite the 2nd Row `(R_2)` as `(R_2-3R_1)`


`[[1,-4,3,-2,9],[0,10,-8,2,-40],[-4,3,-2,1,-4],[-2,1,-4,3,-10]]`


Rewrite the 3rd Row`(R_3)` as`(R_3+4R_1)` 


`[[1,-4,3,-2,9],[0,10,-8,2,-40],[0,-13,10,-7,32],[-2,1,-4,3,-10]]`


Rewrite the 4th Row`(R_4)` as`(R_4+2R_1)`  


`[[1,-4,3,-2,9],[0,10,-8,2,-40],[0,-13,10,-7,32],[0,-7,2,-1,8]]`


Rewrite the 2nd Row`(R_2)` as`(2(R_2+R_3)-R_4)`


`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,-13,10,-7,32],[0,-7,2,-1,8]]`


Rewrite the 3rd Row`(R_3)` as`(R_3+13R_2)`


`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,36,-124,-280],[0,-7,2,-1,8]]`


Rewrite the 4th Row`(R_4)` as `(R_4+7R_2)`  


`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,36,-124,-280],[0,0,16,-64,-160]]`


Rewrite the 3rd Row`(R_3)` as `(R_3-R_4)`


`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,20,-60,-120],[0,0,16,-64,-160]]`


Rewrite the 3rd Row by dividing it with 20,


`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,1,-3,-6],[0,0,16,-64,-160]]`


Rewrite the 4th Row by dividing it with 16,


`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,1,-3,-6],[0,0,1,-4,-10]]`


Rewrite the 4th Row as `(R_3-R_4)`


`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,1,-3,-6],[0,0,0,1,4]]`


Now the matrix is in row-echelon form, and we can perform the back substitution on the corresponding system,


`x-4y+3z-2w=9`     -----  Eq:1


`y+2z-9w=-24`        -----  Eq:2


`z-3w=-6`                 -----  Eq:3


`w=4`  


Substitute back the value of w in Eq:3,


`z-3(4)=-6`


`z-12=-6` 


`z=-6+12`


`z=6`


Substitute back the value of w and z in Eq:2,


`y+2(6)-9(4)=-24`


`y+12-36=-24`


`y=-24+36-12`


`y=0`


Substitute back the value of w,z and y in Eq:1,


`x-4y+3z-2w=9`


`x-4(0)+3(6)-2(4)=9`


`x+18-8=9`


`x=9+8-18`


`x=-1`


So the solutions are x=-1,y=0,z=6 and w=4

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