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2y − 3z = 0 X + 3y = − 4 3x + 4y = 3 - Mathematics

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प्रश्न

2y − 3z = 0
x + 3y = − 4
3x + 4y = 3

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उत्तर

These equations can be written as
0x + 2y − 3z = 0
x + 3y + 0z = − 4
 3x + 4y + 0z = 3

\[D = \begin{vmatrix}0 & 2 & - 3 \\ 1 & 3 & 0 \\ 3 & 4 & 0\end{vmatrix}\] 
\[ = 0(0 - 0) - 2(0 - 0) - 3(4 - 9)\] 
\[ = 15\] 
\[ D_1 = \begin{vmatrix}0 & 2 & - 3 \\ - 4 & 3 & 0 \\ 3 & 4 & 0\end{vmatrix}\] 
\[ = 0(0 - 0) - 2(0 - 0) - 3( - 16 - 9)\] 
\[ = 75\] 
\[ D_2 = \begin{vmatrix}0 & 0 & - 3 \\ 1 & - 4 & 0 \\ 3 & 3 & 0\end{vmatrix}\] 
\[ = 0(0 - 0) - 0(0 - 0) - 3(3 + 12)\] 
\[ = - 45\] 
\[ D_3 = \begin{vmatrix}0 & 2 & 0 \\ 1 & 3 & - 4 \\ 3 & 4 & 3\end{vmatrix}\] 
\[ = 0(9 + 16) - 2(3 + 12) - 0(4 - 9)\] 
\[ = - 30\] 
Now,
\[x = \frac{D_1}{D} = \frac{75}{15} = 5\] 
\[y = \frac{D_2}{D} = \frac{- 45}{15} = - 3\] 
\[z = \frac{D_3}{D} = \frac{- 30}{15} = - 2\] 
\[ \therefore x = 5, y = - 3\text{ and }z = - 2\]

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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 15 | पृष्ठ ८४

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