X − 4y − Z = 11 2x − 5y + 2z = 39 − 3x + 2y + Z = 1 - Mathematics

प्रश्न

x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1

उत्तर

Given: x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1

$D = | \begin{matrix} 1 & - 4 & - 1 \\ 2 & - 5 & 2 \\ - 3 & 2 & 1 \end{matrix} |$
$= 1 (- 5 - 4) - (- 4) (2 + 6) + (- 1) (4 - 15)$
$= 1 (- 9) - (- 4) (8) + (- 1) (- 11) = 34$
$D_{1} = | \begin{matrix} 11 & - 4 & - 1 \\ 39 & - 5 & 2 \\ 1 & 2 & 1 \end{matrix} |$
$= 11 (- 5 - 4) - (- 4) (39 - 2) + (- 1) (78 + 5)$
$= 11 (- 9) - (- 4) (37) + (- 1) (83) = - 34$
$D_{2} = | \begin{matrix} 1 & 11 & - 1 \\ 2 & 39 & 2 \\ - 3 & 1 & 1 \end{matrix} |$
$= 1 (39 - 2) - 11 (2 + 6) + (- 1) (2 + 117)$
$= 1 (37) - 11 (8) + (- 1) (119) = - 170$
$D_{3} = | \begin{matrix} 1 & - 4 & 11 \\ 2 & - 5 & 39 \\ - 3 & 2 & 1 \end{matrix} |$
$= 1 (- 5 - 78) - (- 4) (2 + 117) + 11 (4 - 15)$
$= 1 (- 83) - (- 4) (119) + 11 (- 11) = 272$
Now,
$x = \frac{D_{1}}{D} = \frac{- 34}{34} = - 1$
$y = \frac{D_{2}}{D} = \frac{- 170}{34} = - 5$
$z = \frac{D_{3}}{D} = \frac{272}{34} = 8$
$∴ x = - 1, y = - 5 and z = 8$

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Applications of Determinants and Matrices

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Q 12 Q 11 Q 13

पाठ 6: Determinants - Exercise 6.4 [पृष्ठ ८४]

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पाठ 6 Determinants
Exercise 6.4 | Q 12 | पृष्ठ ८४

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X − 4y − Z = 11 2x − 5y + 2z = 39 − 3x + 2y + Z = 1 - Mathematics

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X − 4y − Z = 11 2x − 5y + 2z = 39 − 3x + 2y + Z = 1 - Mathematics

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