मराठी

​Solve the Following Determinant Equation: ∣ ∣ ∣ ∣ 1 1 X P + 1 P + 1 P + X 3 X + 1 X + 2 ∣ ∣ ∣ ∣ = 0

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प्रश्न

​Solve the following determinant equation:

\[\begin{vmatrix}1 & 1 & x \\ p + 1 & p + 1 & p + x \\ 3 & x + 1 & x + 2\end{vmatrix} = 0\]
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उत्तर

\[\text{ Let }∆ = \begin{vmatrix}1 & 1 & x \\ p + 1 & p + 1 & p + x \\ 3 & x + 1 & x + 2\end{vmatrix}\] 

\[ = \begin{vmatrix}1 & 1 & x \\ p & p & p \\ 3 & x + 1 & x + 2\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\] 

\[ = p\begin{vmatrix}1 & 1 & x \\ 1 & 1 & 1 \\ 3 & x + 1 & x + 2\end{vmatrix}\] 

\[ = p\begin{vmatrix}1 & 1 & x \\ 1 & 1 & 1 \\ 2 & x & 2\end{vmatrix} \left[\text{ Applying }R_3 \to R_3 - R_1 \right]\] 

\[ = p\begin{vmatrix}0 & 1 & x \\ 0 & 1 & 1 \\ 2 - x & x & 2\end{vmatrix} \left[\text{ Applying }C_1\to C_1 - C_2 \right]\] 

\[ = p\left\{ \left( 2 - x \right) \times \begin{vmatrix}1 & x \\ 1 & 1\end{vmatrix} \right\} \left[\text{ Expanding along }C_1 \right]\] 

\[ = p\left( 2 - x \right)\left( 1 - x \right) = 0\] 

\[x = 1, 2\]

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पाठ 5: Determinants - Exercise 6.2 [पृष्ठ ६१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 5 Determinants
Exercise 6.2 | Q 52.8 | पृष्ठ ६१
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