Product of all real values of ' b ' such that there is no solution to the system of equations 2 x+5 y+ z=19,-4 x+ b y+6 z=-42,-3 y-b z=81 is

Question

Product of all real values of ' b ' such that there is no solution to the system of equations 2 x+5 y+ z=19,-4 x+ b y+6 z=-42,-3 y-b z=81 is (A) -30 (B) -48 (C) -24 (D) -18

Tardigrade Admin · Accepted Answer

For no solution of given linear Equations value of given determinant is zero by Cramier's rule,
So,

D = ∣ ∣ 2 - 4 0 5 b - 3 16 - b ∣ ∣ = 0

\Rightarrow 2 (- b^{2} + 18) - 5 (4 b) + 1 (12) = 0

\Rightarrow - 2 b^{2} + 36 - 20 b + 12 = 0

\Rightarrow - 2 b^{2} - 20 b + 48 = 0

\Rightarrow b^{2} + 20 b - 24 = 0

Product of roots of above equations is,
Product

= \frac{c}{a} = \frac{- 24}{1} = - 24

Q. Product of all real values of ' b ' such that there is no solution to the system of equations 2x+5y+z=19,−4x+by+6z=−42,−3y−bz=81 is

-30

-48

-24

-18

Q. Product of all real values of ' $b$ ' such that there is no solution to the system of equations $2 x + 5 y + z = 19, - 4 x + b y + 6 z = - 42, - 3 y - b z = 81$ is