Find area of the triangle with vertices at the point given in each of the following: $(2,7),(1,1),(10,8)$
$\frac{43}{2}$ square units
$\frac{45}{2}$ square units
$\frac{49}{2}$ square units
$\frac{47}{2}$ square units
Let $S$ be the set of all $\lambda \in \mathrm{R}$ for which the system of linear equations
$2 x-y+2 z=2$
$x-2 y+\lambda z=-4$
$x+\lambda y+z=4$
has no solution. Then the set $S$
If ${A_\lambda } = \left( {\begin{array}{*{20}{c}}
\lambda &{\lambda - 1}\\
{\lambda - 1}&\lambda
\end{array}} \right);\,\lambda \in N$ then $|A_1| + |A_2| + ..... + |A_{300}|$ is equal to
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{{\omega ^2}}&\omega \\1&\omega &{{\omega ^2}}\end{array}\,} \right| = $
If $\left| {\begin{array}{*{20}{c}}
{a - b - c}&{2a}&{2a}\\
{2b}&{b - c - a}&{2b}\\
{2c}&{2c}&{c - a - b}
\end{array}} \right|$ $ = \left( {a + b + c} \right)\,{\left( {x + a + b + c} \right)^2}$ , $x \ne 0$ and $a + b + c \ne 0$, then $x$ is equal to
The value of the determinant$\left| {\,\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}\,} \right|$is equal to