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

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प्रश्न

2x − y + z = 0
3x + 2y − z = 0
x + 4y + 3z = 0

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उत्तर

 The given system of homogeneous equations can be written in matrix form as follows:
\[\begin{bmatrix}2 & - 1 & 1 \\ 3 & 2 & - 1 \\ 1 & 4 & 3\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[or, AX = O\]
\[\text{ where, }A = \begin{bmatrix}2 & - 1 & 1 \\ 3 & 2 & - 1 \\ 1 & 4 & 3\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }O = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[ \left| A \right| = \begin{vmatrix}2 & - 1 & 1 \\ 3 & 2 & - 1 \\ 1 & 4 & 3\end{vmatrix}\]
\[ = 2\left( 6 + 4 \right) + 1\left( 9 + 1 \right) + 1(12 - 2)\]
\[ = 40\]
\[ \therefore \left| A \right| \neq0\]
So, the given system has only trivial solution, which is given below:
\[x=y=z=0\]

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पाठ 8: Solution of Simultaneous Linear Equations - Exercise 8.2 [पृष्ठ २०]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 8 Solution of Simultaneous Linear Equations
Exercise 8.2 | Q 1 | पृष्ठ २०

संबंधित प्रश्‍न

\[∆ = \begin{vmatrix}\cos \alpha \cos \beta & \cos \alpha \sin \beta & - \sin \alpha \\ - \sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{vmatrix}\]


Find the value of x, if

\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}1^2 & 2^2 & 3^2 & 4^2 \\ 2^2 & 3^2 & 4^2 & 5^2 \\ 3^2 & 4^2 & 5^2 & 6^2 \\ 4^2 & 5^2 & 6^2 & 7^2\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}\left( 2^x + 2^{- x} \right)^2 & \left( 2^x - 2^{- x} \right)^2 & 1 \\ \left( 3^x + 3^{- x} \right)^2 & \left( 3^x - 3^{- x} \right)^2 & 1 \\ \left( 4^x + 4^{- x} \right)^2 & \left( 4^x - 4^{- x} \right)^2 & 1\end{vmatrix}\]


Prove that:

`[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)] = a^3 + b^3 + c^3 -3abc`


\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]


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\[\begin{vmatrix}x + 1 & 3 & 5 \\ 2 & x + 2 & 5 \\ 2 & 3 & x + 4\end{vmatrix} = 0\]

 


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\[\begin{vmatrix}b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c\end{vmatrix} = 3abc - a^3 - b - c^3\]

 


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\[\begin{vmatrix}a + b & b + c & c + a \\ b + c & c + a & a + b \\ c + a & a + b & b + c\end{vmatrix} = 2\begin{vmatrix}a & b & c \\ b & c & a \\ c & a & b\end{vmatrix}\]

 


Prove that :

\[\begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix} = 2 \left( a + b + c \right)^3\]

 


2x + 3y = 10
x + 6y = 4


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


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


3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1


x − y + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10


An automobile company uses three types of steel S1S2 and S3 for producing three types of cars C1C2and C3. Steel requirements (in tons) for each type of cars are given below : 

  Cars
C1
C2 C3
Steel S1 2 3 4
S2 1 1 2
S3 3 2 1

Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.


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3x + y + z = 0
x − 4y + 3z = 0
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If A and B are non-singular matrices of the same order, write whether AB is singular or non-singular.


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\[\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0\]

 


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2x + 3y − z = 0
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is


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x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
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(b) λ ≠ 5
(c) λ = 5, µ ≠ 13
(d) µ ≠ 13


If \[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\] ,find A–1 and hence solve the system of equations x – 2y = 10, 2x + y + 3z = 8 and –2y + = 7.


Show that  \[\begin{vmatrix}y + z & x & y \\ z + x & z & x \\ x + y & y & z\end{vmatrix} = \left( x + y + z \right) \left( x - z \right)^2\]

 

If A = `[[1,1,1],[0,1,3],[1,-2,1]]` , find A-1Hence, solve the system of equations: 

x +y + z = 6

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Let A = `[(i, -i),(-i, i)], i = sqrt(-1)`. Then, the system of linear equations `A^8[(x),(y)] = [(8),(64)]` has ______.


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