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Solve the Following Determinant Equation: ∣ ∣ ∣ ∣ ∣ 1 X X 2 1 a A 2 1 B B 2 ∣ ∣ ∣ ∣ ∣ = 0 , a ≠ B

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प्रश्न

​Solve the following determinant equation:

\[\begin{vmatrix}1 & x & x^2 \\ 1 & a & a^2 \\ 1 & b & b^2\end{vmatrix} = 0, a \neq b\]

 

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उत्तर

\[\text{ Let }∆ = \begin{vmatrix}1 & x & x^2 \\ 1 & a & a^2 \\ 1 & b & b^2\end{vmatrix}\] 

\[ = \begin{vmatrix}1 & x & x^2 \\ 0 & x - a & x^2 - a^2 \\ 1 & b & b^2\end{vmatrix} \left[\text{ Applying }R_2 \text{ to }R_1 - R_2 \right]\] 

\[ = \begin{vmatrix}1 & x & x^2 \\ 0 & x - a & x^2 - a^2 \\ 0 & x - b & x^2 - b^2\end{vmatrix} \left[\text{ Applying }R_3 \text{ to }R_1 - R_3 \right]\] 

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

\[ ∆ = \left( x - a \right)\left( x - b \right)\left( x + b - x - a \right) = 0\] 

\[x = a, b\] 

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Determinants - Exercise 6.2 [पृष्ठ ६१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 5 Determinants
Exercise 6.2 | Q 52.4 | पृष्ठ ६१
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