The system of equations: `x+y+z=5` `x+2y+3z=9` `x+3y+lambdaz=mu` has a unique solution, if `lambda=5,mu=13` (b) `lambda!=5` `lambda=5,mu!=13` (d) `mu!=13`

Explanation

`2 lamda + 3 + 3 - 2 - 9 - lamda != 0`

`lamda != 5`

`/_\_3 = |(1,1,5),(1,2,9),(1,3,mu)| != 0`

`2 mu + 15 + 9 - 10 - 27 - mu != 0`

`mu != 13`

option b is correct