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JEE Main 20206 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2020 · 6 September · Evening Shift · Mathematics · Determinants

The sum of distinct values of λ for which the system of equations :...

The sum of distinct values of λ\lambda for which the system of equations :

\left(\lambda - 1\right)x + \left(3\lambda + 1\right)y + 2\lambdaz = 0</p><p>(λ−1)x+(4λ−2)y+(λ+3)z=0</p><p>\left(\lambda - 1\right)x + \left(4\lambda - 2\right)y + \left(\lambda + 3\right)z = 0

2x+(3λ+1)y+3(λ−1)z=02x + \left(3\lambda + 1\right)y + 3\left(\lambda - 1\right)z = 0,

Has non-zero solutions, is ....... .

Correct answer

3

Step-by-step solution

\left|\begin{cases}\lambda - 1 & 3\lambda + 1 & 2\lambda \\ \lambda - 1 & 4\lambda - 2 & \lambda + 3 \\ 2 & 3\lambda + 1 & 3 ( \lambda - 1 )\end{cases}\right| = 0

R2→R2−R1{R}_{2} \to {R}_{2} - {R}_{1}

R3→R3−R1{R}_{3} \to {R}_{3} - {R}_{1}

\left|{λ−13λ+12λ0λ−3−λ+33−λ0λ−3\begin{cases}\lambda - 1 & 3\lambda + 1 & 2\lambda \\ 0 & \lambda - 3 & - \lambda + 3 \\ 3 - \lambda & 0 & \lambda - 3\end{cases}\right| = 0</p><p></p><p>{C}_{1} \to {C}_{1} + {C}_{3}</p><p>\left|\begin{cases}3\lambda - 1 & 3\lambda + 1 & 2\lambda \\ 3 - \lambda & \lambda - 3 & 3 - \lambda \\ 0 & 0 & \lambda - 3\end{cases}\right| = 0

⇒(λ−3)2[6λ]=0⇒λ=0,3⇒ {\left(\lambda - 3\right)}^{2}\left[6\lambda\right] = 0 ⇒ \lambda = 0, 3

sum of values of λ=3\lambda = 3

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