If $x = a + 2b$ satisfies the cubic $(a, b\in R)$ $f (x)=$ $\left| {\,\begin{array}{*{20}{c}}{a - x}&b&b\\b&{a - x}&b\\b&b&{a - x}\end{array}\,} \right|$ $= 0$, then its other two roots are

Evaluate the determinants : $\left|\begin{array}{cc}x^{2}-x+1 & x-1 \\ x+1 & x+1\end{array}\right|$

If ${\Delta _r} = \left| {\begin{array}{*{20}{c}}
  r&{2r - 1}&{3r - 2} \\ 
  {\frac{n}{2}}&{n - 1}&a \\ 
  {\frac{1}{2}n\left( {n - 1} \right)}&{{{\left( {n - 1} \right)}^2}}&{\frac{1}{2}\left( {n - 1} \right)\left( {3n - 4} \right)} 
\end{array}} \right|$ then the value of $\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $

Let $\mathrm{A}=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right],$ where $0 \leq \theta \leq 2 \pi$. Then

Find values of $\mathrm{k}$ if area of triangle is $4$ square units and vertices are  $(\mathrm{k}, 0),(4,0),(0,2)$

If the system of equations $x+y+z=6 \,; \,2 x+5 y+\alpha z=\beta  \,; \, x+2 y+3 z=14$ has infinitely many solutions, then $\alpha+\beta$ is equal to.