A solution set of the equations `x+2y+z=1`, `x+3y+4z=k`, `x+5y+10z=k^(2)` is

Options

A. `(1+5lambda,-3lambda,lambda)`
B. `(5lambda-1,1-3lambda,lambda)`
C. `(1+6lambda,-2lambda,lambda)`
D. `(1-6lambda,lambda,lambda)`

Answer

A, B

Explanation

`(a,b)`
The given system of equation is
`x+2y+z=1`…….`(1)`
`x+3y+4z=k`……….`(2)`
`x+5y+10z=k^(2)`………….`(3)`
Subtracting `(1)` from `(2)`, we get `y+3z=k-1`
Subtracting `(2)` from `(3)`, we get `2y+6z=k^(2)-k`
`implies2(k-1)=k^(2)-k`
`impliesk^(2)-3k+2=0`
`impliesk=1,k=2`
For `k=1` ,
`y+3z=0` and `y=-3z`
`impliesx-6z+z=1` (from `(1)`)
`impliesx=1+5z`
One set of solution `(1+5lambda,-3lambda,lambda)`where `lambda` is a variable parameter.
For `k=2`
`y+3z=1` and `y=1-3z`
`x+2-6z+z=1` (from`(1)`)
`impliesx=5z-1`
`:.` Another set of solution `(5lambda-1,1-3lambda,lambda)` where `lambda` is a variable parameter.