Evaluate $\Delta=\left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right|$
Expanding along $\mathrm{R}_{1},$ we get
$\Delta {\text{ }} = 0\left| {\begin{array}{*{20}{c}}
0&{\sin \beta } \\
{ - \sin \beta }&0
\end{array}} \right| - \sin \alpha \left| {\begin{array}{*{20}{c}}
{ - \sin \alpha }&{\sin \beta } \\
{\cos \alpha }&0
\end{array}} \right| - \cos \alpha \left| {\begin{array}{*{20}{c}}
{ - \sin \alpha }&0 \\
{\cos \alpha }&{ - \sin \beta }
\end{array}} \right|$
$=0-\sin \alpha(0-\sin \beta \cos \alpha)-\cos \alpha(\sin \alpha \sin \beta-0)$
$=\sin \alpha \sin \beta \cos \alpha-\cos \alpha \sin \alpha \sin \beta=0$
The value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&a&{b + c}\\1&b&{c + a}\\1&c&{a + b}\end{array}\,} \right|$is
The number of real values $\lambda$, such that the system of linear equations $2 x-3 y+5 z=9$ ; $x+3 y-z=-18$ ; $3 x-y+\left(\lambda^{2}-1 \lambda \mid\right) z=16$ has no solution, is :-
An ordered pair $(\alpha , \beta )$ for which the system of linear equations
$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x + \beta y + 2z = 2$ has a unique solution, is
If for some $\alpha$ and $\beta$ in $R,$ the intersection of the following three planes $x+4 y-2 z=1$ ; $x+7 y-5 z=\beta$ ; $x+5 y+\alpha z=5$ is a line in $\mathrm{R}^{3},$ then $\alpha+\beta$ is equal to
Consider the system of linear equations ${a_1}x + {b_1}y + {c_1}z + {d_1} = 0$, ${a_2}x + {b_2}y + {c_2}z + {d_2} = 0$ and ${a_3}x + {b_3}y + {c_3}z + {d_3} = 0$. Let us denote by $\Delta (a,b,c)$ the determinant $\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ if $\Delta (a,b,c) \ne 0$, then the value of $x$ in the unique solution of the above equations is