JEE Advanced 2021Paper 1
- Exam
- JEE Advanced
- Year
- 2021
- Paper
- Paper 1
- Subject
- Mathematics
- Chapter
- Determinants
JEE Advanced 2021 · Paper 1 · Mathematics · Determinants
Let α, β and γ be real numbers such that the system of linear...
Let \gamma|M|represent the determinant of the matrix \\ \\ M=\left[\right] \\ \\ LetP(\alpha, \beta, \gamma)D(0,1,0)P| M |$ is
Correct answer
0
Step-by-step solution
x + 2y + 3z = \alpha4x + 5y + 6z = \beta7x + 8y + 9z = \gamma - 1\Delta = \left|\right| = 0{L}_{1} + \lambda{L}_{2} = {L}_{3}⇒ \left(x + 2y + 3z - \alpha\right) + \lambda ( 4x + 5y + 6z - \beta ) = 7x + 8y + 9z - \gamma + 1\dfrac{1 + 4\lambda}{7} = \dfrac{2 + 5\lambda}{8} = \dfrac{3 + 6\lambda}{9} = \dfrac{\alpha + \lambda\beta}{\gamma - 1}\dfrac{1 + 4\lambda}{7} = \dfrac{2 + 5\lambda}{8} ⇒ \lambda = - 2\dfrac{1 + 4\lambda}{7} = \dfrac{\alpha + \lambdaB}{\gamma - 1}⇒ - 1 = \dfrac{\alpha - 2\beta}{r - 1}⇒ \alpha - 2\beta + \gamma = 1</p><p>Now, \left|M\right| = \left|\begin{cases}\alpha & 2 & \gamma \\ \beta & 1 & 0 \\ - 1 & 0 & 1\end{cases}\right|