Statement $-1$ : The system of linear equations
$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$
$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$
$x - \left( {\sin \,\alpha } \right)y - \left( {\cos \alpha } \right)z = 0$
has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$
Statement $-2$ : The equation in $\alpha $
$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$
has only one solution lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$
Statement $- 1$ is true, Statement $-2$ is true,Statement $-2$ is not correct explantion for Statement $-1$
Statement $-1$ is true, Statement $-2$ is true,Statement $-2$ is a correct explantion for Statement $-1$
Statement $- 1$ is true, Statement $-2$ is false
Statememt $-1$ is false, Statement $-2$ is true.
If $\alpha , \beta \, and \, \gamma$ are real numbers , then $D = \left|{\begin{array}{*{20}{c}}1&{\cos \,(\beta \, - \,\alpha )}&{\cos \,(\gamma \, - \,\alpha )}\\{\cos \,(\alpha \, - \,\beta )}&1&{\cos \,(\gamma \, - \,\beta )}\\{\cos \,(\alpha \, - \,\gamma )}&{\cos \,(\beta \, - \,\gamma )}&1 \end{array}} \right|$ =
$\Delta = \left| {\,\begin{array}{*{20}{c}}{a + x}&b&c\\b&{x + c}&a\\c&a&{x + b}\end{array}\,} \right|$,which of the following is a factor for the above determinant
If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\{2{x^2} + 3x - 1}&{3x}&{3x - 3}\\{{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}\,} \right| = Ax - 12$, then the value of $A $ is
The value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&2&3\\3&5&7\\8&{14}&{20}\end{array}\,} \right|$is
If $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right],$ then show that $|2 A|=4|A|$.