मराठी

X + Y = 1 X + Z = − 6 X − Y − 2z = 3

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प्रश्न

x + y = 1
x + z = − 6
x − y − 2z = 3

बेरीज
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उत्तर

These equations can be written as
xy + 0z = 1
x + 0y + z = − 6
x − y − 2z = 3

\[D = \begin{vmatrix}1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & - 1 & - 2\end{vmatrix}\]

\[ = 1(0 + 1) - 1( - 2 - 1) + 0( - 1 - 0)\]

\[ = 4\]

\[ D_1 = \begin{vmatrix}1 & 1 & 0 \\ - 6 & 0 & 1 \\ 3 & - 1 & - 2\end{vmatrix}\]

\[ = 1(0 + 1) - 1(12 - 3) + 0(6 - 0)\]

\[ = - 8\]

\[ D_2 = \begin{vmatrix}1 & 1 & 0 \\ 1 & - 6 & 1 \\ 1 & 3 & - 2\end{vmatrix}\]

\[ = 1(12 - 3) - 1( - 2 - 1) + 0(3 + 6)\]

\[ = 12\]

\[ D_3 = \begin{vmatrix}1 & 1 & 1 \\ 1 & 0 & - 6 \\ 1 & - 1 & 3\end{vmatrix}\]

\[ = 1(0 - 6) - 1(3 + 6) + 1( - 1 - 0)\]

\[ = - 16\]

\[ \text{ Now } , \]

\[x = \frac{D_1}{D} = \frac{- 8}{4} = - 2\]

\[y = \frac{D_2}{D} = \frac{12}{4} = 3\]

\[z = \frac{D_3}{D} = \frac{- 16}{4} = - 4\]

\[ \therefore x = - 2, y = 3 \text{ and }  z = - 4\]

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पाठ 5: Determinants - Exercise 6.4 [पृष्ठ ८४]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 5 Determinants
Exercise 6.4 | Q 18 | पृष्ठ ८४

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