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​Solve the Following Determinant Equation: ∣ ∣ ∣ ∣ ∣ 1 X X 3 1 B B 3 1 C C 3 ∣ ∣ ∣ ∣ ∣ = 0 , B ≠ C - Mathematics

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प्रश्न

​Solve the following determinant equation:

\[\begin{vmatrix}1 & x & x^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{vmatrix} = 0, b \neq c\]

 

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उत्तर

\[\text{ Let }∆ = \begin{vmatrix}1 & x & x^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{vmatrix}\] 

\[ = \begin{vmatrix}1 & x & x^3 \\ 0 & b - x & b^3 - x^3 \\ 1 & c & c^3\end{vmatrix} \left[\text{ Applying }R_2 \text{ to }R_2 - R_1 \right]\] 

\[ = \begin{vmatrix}1 & x & x^3 \\ 0 & b - x & b^3 - x^3 \\ 0 & c - x & c^3 - x^3\end{vmatrix} \left[\text{ Applying }R_3 \text{ to }R_3 - R_1 \right]\] 

\[ = \begin{vmatrix}1 & x & x^3 \\ 0 & x - b & x^3 - b^3 \\ 0 & x - c & x^3 - c^3\end{vmatrix}\] 

\[ = \left( x - b \right)\left( x - c \right)\begin{vmatrix}1 & x & x^2 \\ 0 & 1 & x^2 + xb + b^2 \\ 0 & 1 & x^2 + xc + c^2\end{vmatrix} \] 

\[ ∆ = \left( x - b \right)\left( x - c \right)\left( x\left( c - b \right) - b^2 + c^2 \right) = 0\] 

\[x = b, c, - \left( b + c \right)\] 

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अध्याय 6: Determinants - Exercise 6.2 [पृष्ठ ६१]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 6 Determinants
Exercise 6.2 | Q 52.6 | पृष्ठ ६१

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