How do you solve the system #x + 2y -4z = 0#, #2x + 3y + z = 1#, #4x + 7y + lamda*z = mu#?

1 Answer
Feb 23, 2017

Answer:

#x = (28-14mu+2 lambda)/(lambda + 7); y = (-16+9mu-lambda)/(lambda + 7) ;z = (mu-1)/(lambda+7)#

Explanation:

This system can be solved a number of ways, such as substitution and elimination, matrix row operations, etc.

Using Matrix Row Operations we want to get: #((1,0,0,x),(0,1,0,y),(0,0,1,z))#

Since we have a 1 in the top left do the following row operations.
To row 2: #-2R_1 + R_2# and to row 3: #-4R_1 + R_3#

#((1,2,-4,0),(2,3,1,1),(4,7,lambda,mu)) => ((1,2,-4,0),(0,-1,9,1),(0,-1,16+lambda,mu))#

To get a 1 in the 2nd column of row 2: #-R_2#
#((1,2,-4,0),(0,1,-9,-1),(0,-1,16+lambda,mu))#

To get (0, 1, 0) in the 2nd column:
To row 1: #-2R_2+R_1# and to row 3: #R_2 + R_3#
#((1,0,14,2),(0,1,-9,-1),(0,0,7+lambda,mu-1))#

To get a 1 in the 3rd row, 3rd column: #R_3/(lambda+7)#
#((1,0,14,2),(0,1,-9,-1),(0,0,1,(mu-1)/(lambda+7)))#

To get (0,0,1) in the 3rd column:
To row 1: #-14R_3+R_1# and to row 2: #9R_3+R_2#
#((1,0,0,-14(mu-1)/(lambda + 7) +2),(0,1,0,9(mu-1)/(lambda+7)-1),(0,0,1,(mu-1)/(lambda+7))) => ((1,0,0,(28-14mu+2lambda)/(lambda + 7)),(0,1,0,(9mu-lambda-16)/(lambda+7)),(0,0,1,(mu-1)/(lambda+7)))#

So #x = (28-14mu+2 lambda)/(lambda + 7); y = (-16+9mu-lambda)/(lambda + 7) ;z = (mu-1)/(lambda+7)#