मराठी

Evaluate : ∣ ∣ ∣ ∣ X + λ X X X X + λ X X X X + λ ∣ ∣ ∣ ∣

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प्रश्न

Evaluate :

\[\begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]

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उत्तर

\[∆ = \begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]

\[ = \begin{vmatrix}\lambda & 0 & x \\ - \lambda & \lambda & x \\ 0 & - \lambda & x + \lambda\end{vmatrix} \left[\text{ Applying }C_1 \to C_1 - C_2 , C_2 \to C_2 - C_3 \right]\]

\[ = \begin{vmatrix}\lambda & 0 & x \\ - \lambda & 0 & 2x + \lambda \\ 0 & - \lambda & x + \lambda\end{vmatrix} \left[ \text{ Applying }R_1 \text{ to } R_2 + R_3 \right]\]

\[ = \lambda\begin{vmatrix}0 & 2x + \lambda \\ - \lambda & x + \lambda\end{vmatrix} + x\begin{vmatrix}- \lambda & 0 \\ 0 & - \lambda\end{vmatrix}\]

\[ = \lambda[\lambda(2x + \lambda)] + x \lambda^2 \]

\[ = \lambda^2 (2x + \lambda + \lambda^2 x)\]

\[ = 3 \lambda^2 x + \lambda^3 \]

\[ = \lambda^2 (3x + \lambda )\]

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पाठ 5: Determinants - Exercise 6.2 [पृष्ठ ५८]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 5 Determinants
Exercise 6.2 | Q 5 | पृष्ठ ५८

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