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प्रश्न
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उत्तर
Here,
\[\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ - 1 \\ 0\end{bmatrix} \]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ - 1 \\ 0\end{bmatrix}\]
\[ \therefore x = 1, y = - 1\text{ and }z = 0\]
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संबंधित प्रश्न
Examine the consistency of the system of equations.
x + 3y = 5
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Find the value of x, if
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\[\begin{vmatrix}8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3\end{vmatrix}\]
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\[\begin{vmatrix}0 & x & y \\ - x & 0 & z \\ - y & - z & 0\end{vmatrix}\]
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\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix}\]
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\[\begin{vmatrix}x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} = \left( 5x + 4 \right) \left( 4 - x \right)^2\]
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Prove that :
Prove that :
Given: x + 2y = 1
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x − 2y − z = 1
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\[2 \lambda x - 2y + 3z = 0\]
\[ x + \lambda y + 2z = 0\]
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x − y + 2z = 0
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x + y + z = 5
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x + 3y + λz = µ
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(a) λ = 5, µ = 13
(b) λ ≠ 5
(c) λ = 5, µ ≠ 13
(d) µ ≠ 13
Solve the following by inversion method 2x + y = 5, 3x + 5y = −3
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Choose the correct option:
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x + y – 3 = 0
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