Let S be the set of values of , for which the system of equations 6 x-3y+3z=4 ^2, 2x+6 y+4…

Question

Let $S$ be the set of values of $\lambda $, for which the system of equations

$6\lambda x-3y+3z=4{\lambda }^{2}$,

$2x+6\lambda y+4z=1$,

$3x+2y+3\lambda z=\lambda $ has no solution. Then $12\sum _{i\in S}|\lambda |$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 24

Accepted Answer

Given that $S$ be the set of values of $\lambda $ for which given system of equations has no solution.

Therefore for the given set of equations

$\Delta=\begin{vmatrix} 6\lambda  \quad  -3 \quad  3 \ 2 \quad  6\lambda  \quad  4 \ 3 \quad  2 \quad  3\lambda \end{vmatrix}=0$

$\begin{gathered} \Rightarrow 6\lambda (18{\lambda }^{2}-8)+3(6\lambda -12)+3(4-18\lambda )=0 \ \Rightarrow 18{\lambda }^{3}-14\lambda -4=0 \ \Rightarrow (\lambda -1)(3\lambda +1)(3\lambda +2)=0 \ \Rightarrow \lambda =1,-\frac{1}{3},-\frac{2}{3} \end{gathered}$

Also for each values of $\lambda =1,\frac{-1}{3},\frac{-2}{3}$, we have

$\begin{vmatrix} 6\lambda  \quad  -3 \quad  4{\lambda }^{2} \ 2 \quad  6\lambda  \quad  1 \ 3 \quad  2 \quad  \lambda \end{vmatrix}
e 0$

which implies that, for each values of $\lambda $, the given system of equations has no solution.

$\begin{gathered} S\in {1,\frac{-1}{3},\frac{-2}{3}} \ 12\sum _{\lambda \in S}|\lambda, \ =12(|1|+|\frac{-1}{3}|+|\frac{-2}{3}|) \ =12(1+\frac{1}{3}+\frac{2}{3})=12(\frac{6}{3})=24 \end{gathered}$

Solutions · 1