Q100 / 160 · AP EAMCET 2020 (17 Sep Shift 2) · Mathematics
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
- Correct answerC-24
- D-18
Solution
For no solution of given linear Equations value of given determinant is zero by Cramier's rule,
So,
\(\begin{array}{rlrl}
& & D=\left|\begin{array}{ccc}
2 & 5 & 1 \\
-4 & b & 6 \\
0 & -3 & -b
\end{array}\right| & =0 \\
\Rightarrow & 2\left(-b^2+18\right)-5(4 b)+1(12) & =0 \\
\Rightarrow & & -2 b^2+36-20 b+12 & =0 \\
\Rightarrow & & -2 b^2-20 b+48 & =0 \\
\Rightarrow & & b^2+20 b-24 & =0
\end{array}\)
Product of roots of above equations is,
\(\text {Product }=\frac{c}{a}=\frac{-24}{1}=-24\)
Hence, the correct answer is option: 'C'.