Jee Main 2024 01 02 2024 Shift 2 - Question10
Question 10
If system of equation
$x+2 y+3 z=5$
$3 x+3 y+z=9$
$x+4 y+\lambda z=\mu$
have infinitely many solutions then the value of $3 \lambda+\mu$ equals to
(1) 17
(2) 21
(3) 43
(4) 34
Show Answer
Answer: (4)
Solution:
$\Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 3 & 3 & 1 \\ 1 & 4 & \lambda\end{array}\right|=0$
$(3 \lambda-4)-2(3 \lambda-1)+3(12-3)=0$
$3 \lambda-4-6 \lambda+2+27=0$
$-3 \lambda+25=0$
$\lambda=\frac{25}{3}$
$\Delta _1=\left|\begin{array}{lll}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 4 & \lambda\end{array}\right|=0$
$=5(3 \lambda-4)-2(9 \lambda-\mu)+3(36-3 \mu)=0$
For $\lambda=\frac{25}{3} \mu=9$
Now for $\lambda=\frac{25}{3} \& \mu=9 \Delta=\Delta _1=\Delta _2=\Delta _3=0$
$3 \lambda+\mu=\frac{25}{3} \times 3+9=25+9=34$
BETA
