Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 399 Question 399 of 472 Medium MCQ
If \[\alpha\] + \[\beta\] + \[\gamma\] = 2\[\pi\], then the system of equations
x + (cos \[\gamma\])y + (cos \[\beta\])z = 0
(cos \[\gamma\])x + y + (cos \[\alpha\])z = 0
(cos \[\beta\])x + (cos \[\alpha\])y + z = 0
has :
Solutions · 1 Post your solution Official solution
Given \[\alpha\] + \[\beta\] + \[\gamma\] = 2\[\pi\]
\[\Delta = \left| { ( 1 cos γ cos β cos γ 1 cos α cos β cos α 1 ) \begin{pmatrix} 1 & {\cos \gamma } & {\cos \beta } \\
{\cos \gamma } & 1 & {\cos \alpha } \\
{\cos \beta } & {\cos \alpha } & 1 \\
\end{pmatrix} 1 cos γ cos β cos γ 1 cos α cos β cos α 1 } \right|\]
\[ = 1 - {\cos ^2}\alpha - \cos \gamma (\cos \gamma - \cos \alpha \cos \beta ) + \cos \beta (\cos \alpha \cos \gamma - \cos \beta )\]
\[ = 1 - {\cos ^2}\alpha - {\cos ^2}\beta - {\cos ^2}\gamma + 2\cos \alpha \cos \beta \cos \gamma \]
\[ = {\sin ^2}\alpha - {\cos ^2}\beta - \cos \gamma (\cos \gamma - 2\cos \alpha \cos \beta )\]
\[ = - \cos (\alpha + \beta )\cos (\alpha - \beta ) - \cos \gamma (\cos (2\pi - (\alpha - \beta )) - 2\cos \alpha \cos \beta )\]
\[ = - \cos (2\pi - \gamma )\cos (\alpha - \beta ) - \cos \gamma (\cos (\alpha + \beta ) - 2\cos \alpha \cos \beta )\]
\[ = - \cos \gamma \cos (\alpha - \beta ) + \cos \gamma (\cos \alpha \cos \beta + \sin \alpha \sin \beta )\]
\[ = - \cos \gamma \cos (\alpha - \beta ) + \cos \gamma \cos (\alpha - \beta )\]
= 0 = 0 = 0
So, the system of equation has infinitely many solutions. Answer the question to unlock