Consider the system of linear equation x+y+z= 4,\ x+2y+2λ z=10,\ x+3y+4λ² z=²+15 where λ, R. Which one of the following statements is NOT correct? [30-Jan-2024 Shift 1]

Correct Answer is : The system is inconsistent if λ=12λ=1/2λ=21 and μ≠1≠ 1μ=1

The system has unique solution if λ≠ 1/2 and ≠ 1,15

The system is inconsistent if λ=1/2 and ≠ 1

The system has infinite number of solutions if λ=1/2 and =15

The system is consistent if λ≠ 1/2

Correct Answer is : The system is inconsistent if λ=12λ=1/2λ=21 and μ≠1≠ 1μ=1

Test Index

JEE Main 30 Jan 2024 Shift 1 Paper

Section: Mathematics

Question : 13 of 90

Marks: +1, -0

x+y+z=

4\mu,\ x+2y+2\lambda z=10\mu,\ x+3y+4\lambda^2 z=\mu^2+15

\lambda,\mu\in R

The system has unique solution if $\lambda\neq \frac{1}{2}$ and $\mu\neq 1,15$
The system is inconsistent if $\lambda=\frac{1}{2}$ and $\mu\neq 1$
The system has infinite number of solutions if $\lambda=\frac{1}{2}$ and $\mu=15$
The system is consistent if $\lambda\neq \frac{1}{2}$

Solution: 👈: Video Solution

\;x+y+z=4\mu,\ x+2y+2\lambda z=10\mu,

x+3y+4\lambda^2 z=\mu^2+15

\;\Delta=\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix}

=(2\lambda-1)^2

For unique solution

\Delta\neq 0,\ 2\lambda-1\neq 0,\ \left(\lambda\neq \frac{1}{2}\right)

Let

\Delta=0,\ \lambda=\frac{1}{2}

\;\Delta_y=0,\ \Delta_x=\Delta_z

=\begin{vmatrix} 4\mu & 1 & 1 \\ 10\mu & 2 & 1 \\ \mu^2+15 & 3 & 1 \end{vmatrix}

\;=(\mu-15)(\mu-1)

For infinite solution

\lambda=\frac{1}{2},\ \mu=1

or 15

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