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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 α,βand\alpha, \beta and\gammaberealnumberssuchthatthesystemoflinearequations<br>be real numbers such that the system of linear equations<br> \\x+2y+3z=α4x+5y+6z=β7x+8y+9z=γ−1\begin{array}{c}\\ x+2 y+3 z=\alpha\\ \\ 4 x+5 y+6 z=\beta\\ \\ 7 x+8 y+9 z=\gamma-1 \\ \end{array}isconsistent.Let\\ \\ is consistent. Let|M|represent the determinant of the matrix \\ \\ M=\left[α2γβ10−101\begin{array}{ccc}\\ \alpha& 2 & \gamma\\ \\ \beta& 1 & 0 \\ \\ -1 & 0 & 1 \\ \end{array}\right] \\ \\ LetPbetheplanecontainingallthosebe the plane containing all those(\alpha, \beta, \gamma)forwhichtheabovesystemoflinearequationsisconsistent,andfor which the above system of linear equations is consistent, andDbethesquareofthedistanceofthepointbe the square of the distance of the point(0,1,0)fromtheplanefrom the planeP.<br><p>Thevalueof.<br><p>The value of| M |$ is

Correct answer

0

Step-by-step solution

x + 2y + 3z = \alpha</p><p></p><p>4x + 5y + 6z = \beta</p><p></p><p>7x + 8y + 9z = \gamma - 1</p><p>Thesystemoflinearequationsisconsistentitmeansithasuniquesolutionorinfinitesolutions</p><p>Here,</p><p>The system of linear equations is consistent it means it has unique solution or infinite solutions</p><p>Here,\Delta = \left|{123456789\begin{cases}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{cases}\right| = 0</p><p>Hence,thereareinfinitelymanysolutions.</p><p>Ifequationshaveinfinitelymanysolutions,thentheequationsarelinearlyconnected,i.e.,</p><p>Hence, there are infinitely many solutions.</p><p>If equations have infinitely many solutions, then the equations are linearly connected, i.e.,{L}_{1} + \lambda{L}_{2} = {L}_{3}</p><p></p><p>⇒ \left(x + 2y + 3z - \alpha\right) + \lambda ( 4x + 5y + 6z - \beta ) = 7x + 8y + 9z - \gamma + 1</p><p></p><p>\dfrac{1 + 4\lambda}{7} = \dfrac{2 + 5\lambda}{8} = \dfrac{3 + 6\lambda}{9} = \dfrac{\alpha + \lambda\beta}{\gamma - 1}</p><p></p><p>\dfrac{1 + 4\lambda}{7} = \dfrac{2 + 5\lambda}{8} ⇒ \lambda = - 2</p><p>Also,</p><p>Also,\dfrac{1 + 4\lambda}{7} = \dfrac{\alpha + \lambdaB}{\gamma - 1}</p><p></p><p>⇒ - 1 = \dfrac{\alpha - 2\beta}{r - 1}</p><p></p><p>⇒ \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|

=α+2(−β)+γ(1)=α−2β+γ=1= \alpha + 2 ( - \beta ) + \gamma ( 1 ) = \alpha - 2\beta + \gamma = 1

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