Statement -1 : The system of linear equations | vedclass

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Question

${\Delta _1} = \left| {\begin{array}{*{20}{c}}
1&{\sin \alpha }&{\cos \alpha }\
1&{\cos \alpha }&{\sin \alpha }\
1&{ - \sin \

Vedclass · Accepted Answer

Statement $- 1$ is true, Statement $-2$ is false

Vedclass · Answer

${\Delta _1} = \left| {\begin{array}{*{20}{c}}
1&{\sin \alpha }&{\cos \alpha }\
1&{\cos \alpha }&{\sin \alpha }\
1&{ - \sin \alpha }&{\cos \alpha }
\end{array}} ight|$

$ = \left| {\begin{array}{*{20}{c}}
0&{\sin \alpha  - \cos \alpha }&{\cos \alpha  - \sin \alpha }\
0&{\cos \alpha  + \sin \alpha }&{\sin \alpha  - \cos \alpha }\
1&{ - \sin \alpha }&{\cos \alpha }
\end{array}} ight|$

$ = {\left( {\sin \alpha  - \cos \alpha } ight)^2} - \left( {{{\cos }^2}\alpha  - {{\sin }^2}\alpha } ight)$

$ = {\sin ^2}\alpha  - {\cos ^2}\alpha  - 2\sin \alpha .\cos \alpha  - {\cos ^2}\alpha  + {\sin ^2}\alpha $

$ = 2{\sin ^2}\alpha  - 2\sin \alpha .\cos \alpha $

$ = 2\sin \alpha \left( {\sin \alpha  - \cos \alpha } ight)$

Now, $\sin \alpha  - \cos \alpha  = 0$ for only

$\alpha  = \frac{\pi }{4}$ in $\left( {0,\frac{\pi }{2}} ight)$

${\Delta _1} = 2\left( {\sin \alpha } ight) 	imes 0 = 0$

Since value of $\sin \alpha $ is finite for $\alpha  \in \left( {0,\frac{\pi }{2}} ight)$

Hence non- trivivial solution for only one value of $\alpha $ in $\left( {0,\frac{\pi }{2}} ight)$

$\left| {\begin{array}{*{20}{c}}
{\cos \alpha }&{\sin \alpha }&{\cos \alpha }\
{\sin \alpha }&{\cos \alpha }&{\sin \alpha }\
{\cos \alpha }&{ - \sin \alpha }&{ - \cos \alpha }
\end{array}} ight| = 0$

$ \Rightarrow \left| {\begin{array}{*{20}{c}}
0&{\sin \alpha }&{\cos \alpha }\
0&{\cos \alpha }&{\sin \alpha }\
{2\cos \alpha }&{ - \sin \alpha }&{ - \cos \alpha }
\end{array}} ight| = 0$

$ \Rightarrow 2\cos \alpha \left( {{{\sin }^2}\alpha  - {{\cos }^2}\alpha } ight) = 0$

$	herefore \cos \alpha  = 0$ or ${\sin ^2}\alpha  - {\cos ^2}\alpha  = 0$

But $\cos \alpha  = 0$ not possible for any value of 

$\alpha  \in \left( {0,\frac{\pi }{2}} ight)$

$	herefore {\sin ^2}\alpha  - {\cos ^2}\alpha  = 0 \Rightarrow \sin \alpha  =  - \cos \alpha $

which is also not possible for any value of 

$\alpha  \in \left( {0,\frac{\pi }{2}} ight)$

Hence, there is no solution.

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