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∣ ∣ ∣ ∣ 1 + a 1 1 1 1 + a A 1 1 1 + a ∣ ∣ ∣ ∣ = a 3 + 3 a 2

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प्रश्न

\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]

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उत्तर

\[∆ = \begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & 1 \\ 1 & 1 & 1 + a\end{vmatrix}\]

\[ = 1 + a \begin{vmatrix}1 + a & 1 \\ 1 & 1 + a\end{vmatrix} - 1\begin{vmatrix}1 & 1 \\ 1 & 1 + a\end{vmatrix} + 1\begin{vmatrix}1 & 1 + a \\ 1 & 1\end{vmatrix} \left[\text{ Expanding }\right]\]

\[ = (1 + a)\left[ (1 + a )^2 - 1 \right] - 1(1 + a - 1) + (1 - 1 - a)\]

\[ = (1 + a)[1 + a^2 + 2a - 1] - a - a\]

\[ = 1 + a + a^2 + a^3 + 2a + 2 a^2 - 2a\]

\[ = a^3 + 3 a^2\]

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अध्याय 5: Determinants - Exercise 6.2 [पृष्ठ ६१]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 5 Determinants
Exercise 6.2 | Q 41 | पृष्ठ ६१

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