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उत्तर
`=>|(15-2x-14+2x,11-3x,7-x),(11-28,17,14),(10-26,16,13)|=0` `["Applying" C_1->C_1-2C_3]`
`=>|(1,11-3x,7-x),(-17,17,14),(-16,16,13)|=0`
`=>|(12-3x,4-2x,7-x),(0,3,14),(0,3,13)|=0` `["Applying" C_1->C_1+C_2 "and" C_2->C_2-C_3]`
\[ \Rightarrow \left( 12 - 3x \right)\left( \left( 3 \times 13 \right) - \left( 3 \times 14 \right) \right) = 0\]
\[ \Rightarrow \left( 12 - 3x \right)\left( - 3 \right) = 0\]
\[ \Rightarrow 12 - 3x = 0\]
\[ \Rightarrow 3x = 12\]
\[ \Rightarrow x = 4\]
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संबंधित प्रश्न
Examine the consistency of the system of equations.
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Evaluate
\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2\end{vmatrix}\]
Prove the following identities:
\[\begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix} = \left( 5x + \lambda \right) \left( \lambda - x \right)^2\]
Using properties of determinants prove that
\[\begin{vmatrix}x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} = \left( 5x + 4 \right) \left( 4 - x \right)^2\]
\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]
Without expanding, prove that
\[\begin{vmatrix}a & b & c \\ x & y & z \\ p & q & r\end{vmatrix} = \begin{vmatrix}x & y & z \\ p & q & r \\ a & b & c\end{vmatrix} = \begin{vmatrix}y & b & q \\ x & a & p \\ z & c & r\end{vmatrix}\]
Solve the following determinant equation:
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(2, 7), (1, 1) and (10, 8)
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Solve each of the following system of homogeneous linear equations.
2x + 3y + 4z = 0
x + y + z = 0
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Find the value of the determinant \[\begin{vmatrix}2^2 & 2^3 & 2^4 \\ 2^3 & 2^4 & 2^5 \\ 2^4 & 2^5 & 2^6\end{vmatrix}\].
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Let \[\begin{vmatrix}x & 2 & x \\ x^2 & x & 6 \\ x & x & 6\end{vmatrix} = a x^4 + b x^3 + c x^2 + dx + e\]
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3x + 4y + 7z = 14
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A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.
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3x – 2y – kz = 10
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