SOLUTION: solve the system: x1+x2 =5 x2+x3 =5 x3+x4=-5 x1 +x4=-5 [x1] = [ ] + [ ]s [x2] = [ ] + [ ]s [x3] = [ ] + [ ]s [x4] = [ ] + [ ]s

Algebra ->  Matrices-and-determiminant -> SOLUTION: solve the system: x1+x2 =5 x2+x3 =5 x3+x4=-5 x1 +x4=-5 [x1] = [ ] + [ ]s [x2] = [ ] + [ ]s [x3] = [ ] + [ ]s [x4] = [ ] + [ ]s      Log On


   



Question 831301: solve the system:
x1+x2 =5
x2+x3 =5
x3+x4=-5
x1 +x4=-5
[x1] = [ ] + [ ]s
[x2] = [ ] + [ ]s
[x3] = [ ] + [ ]s
[x4] = [ ] + [ ]s

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!


Put in missing variables with 0 coefficients
and give single variables 1 coefficents



Put the coefficients in a 4×5 augmented matrix and do row operations
to get 0's below the diagonals:



-1*R1+R4->R4   (Multiply row 1 by -1, add to row 4, put it in row 4.) 



R2+R4->R4    (Add row 2 and R4 and put it in row 4)  



-R3+R4->R4





Eliminate the 0s and 1 coefficients, which
means we eliminate the entire bottom equation. 



We let x%5B4%5D=s



Solve the bottom equation of the system: x%5B3%5D=-5-s

Substitute in the middle equation and solve:

x%5B2%5D%2Bx%5B3%5D+=+5
x%5B2%5D%2B%28-5-s%29+=+5
{{x[2]-5-s = 5}}}
{{x[2]= 10+s}}}

Substitute in the top equation

x%5B1%5D%2Bx%5B2%5D+=+5
x%5B1%5D%2B%2810%2Bs%29+=+5
x%5B1%5D%2B10%2Bs+=+5
x%5B1%5D=-5-s


x%5B1%5D=-5-s
x%5B2%5D=+10%2Bs
x%5B3%5D=-5-s
x%5B4%5D=s

or in the form that you gave:

[x1] = [5] + [-1]s 
[x2] = [10] + [1]s
[x3] = [-5] + [-1]s 
[x4] = [0] + [1]s

Edwin