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3x + Ay = 4 2x + Ay = 2, a ≠ 0 - Mathematics

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प्रश्न

3x + ay = 4
2x + ay = 2, a ≠ 0

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उत्तर

\[\text{ Given} : 3x + ay = 4\] 
\[ 2x + ay = 2 \] 
Using Cramer's rule, we get
\[ D = \begin{vmatrix} 3 & a\\2 & a \end{vmatrix} = 3a - 2a = a\] 
\[ D_1 = \begin{vmatrix} 4 & a\\2 & a \end{vmatrix} = 4a - 2a = 2a\] 
\[ D_2 = \begin{vmatrix} 3 & 4 \\2 & 2 \end{vmatrix} = 6 - 8 = - 2\] 
Now, 
\[x = \frac{D_1}{D} = \frac{2a}{a} = 2\] 
\[y = \frac{D_2}{D} = \frac{- 2}{a} = - \frac{2}{a}\] 
\[ \therefore x = 2\text{ and }y = - \frac{2}{a}\]

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अध्याय 6: Determinants - Exercise 6.4 [पृष्ठ ८४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 6 Determinants
Exercise 6.4 | Q 6 | पृष्ठ ८४

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