Question

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

Answer 1

`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)`