NTA Abhyas JEE Main2020MathematicsDeterminantsPractice
If the system of equations 14 x - 3 y + z = 12 ,   x - 2 y = 0 and x + 2 z = 0 has a solution x 0 , y 0 , z 0 , then the value of x 0 2 + y 0 2 + z 0 2 is equal to
Options
- A3 2
- B3 4
- C9 2
- D9 4
Correct answer
A. 3 2
Step-by-step solution
Using cramer’s rule, we get, ∆ = 14 - 3 1 1 - 2 0 1 0 2 = 14 - 4 + 3 2 + 1 2 = - 48 ∆ 1 = 12 - 3 1 0 - 2 0 0 0 2 = 12 - 4 = - 48 ∆ 2 = 14 12 1 1 0 0 1 0 2 = - 12 2 = - 24 ∆ 3 = 14 - 3 12 1 - 2 0 1 0 0 = 12 2 = 24 ⇒ x = ∆ 1 ∆ = 1 , y = ∆ 2 ∆ = 1 2 , z = ∆ 3 ∆ = - 1 2 ⇒ x 0 2 + y 0 2 + z 0 2 = 1 + 1 4 + 1 4 = 3 2