The system $x+2 y+3 z=4,4 x+5 y+3 z=5,3 x+4 y+3 z=\lambda$ is consistent and $3 \lambda=n+100$, then $n=$

The system $x+2 y+3 z=4,4 x+5 y+3 z=5,3 x+4 y+3 z=\lambda$ is consistent and $3 \lambda=n+100$, then $n=$
  1. -42
  2. -86
  3. 16
  4. -24

Solution

$\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|=3-2(12-9)+3 \times(16-15)=0$
For equation to be consistent $\begin{aligned} & \Delta_3=0 \Rightarrow\left|\begin{array}{lll} 1 & 2 & 4 \\ 4 & 5 & 5 \\ 3 & 4 & \lambda \end{array}\right|=0 \\ & \Rightarrow 5 \lambda-20-2(4 \lambda-15)+4=0 \Rightarrow \lambda=\frac{14}{3} \\ & \Rightarrow 3 \lambda=-14=n+100 \Rightarrow 14=n+100 \Rightarrow n=-86 \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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