Advertisements
Advertisements
प्रश्न
Evaluate :
\[\begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]
Advertisements
उत्तर
\[∆ = \begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]
\[ = \begin{vmatrix}\lambda & 0 & x \\ - \lambda & \lambda & x \\ 0 & - \lambda & x + \lambda\end{vmatrix} \left[\text{ Applying }C_1 \to C_1 - C_2 , C_2 \to C_2 - C_3 \right]\]
\[ = \begin{vmatrix}\lambda & 0 & x \\ - \lambda & 0 & 2x + \lambda \\ 0 & - \lambda & x + \lambda\end{vmatrix} \left[ \text{ Applying }R_1 \text{ to } R_2 + R_3 \right]\]
\[ = \lambda\begin{vmatrix}0 & 2x + \lambda \\ - \lambda & x + \lambda\end{vmatrix} + x\begin{vmatrix}- \lambda & 0 \\ 0 & - \lambda\end{vmatrix}\]
\[ = \lambda[\lambda(2x + \lambda)] + x \lambda^2 \]
\[ = \lambda^2 (2x + \lambda + \lambda^2 x)\]
\[ = 3 \lambda^2 x + \lambda^3 \]
\[ = \lambda^2 (3x + \lambda )\]
