मराठी

Two Institutions Decided to Award Their Employees for the Three Values of Resourcefulness, Competence and Determination in the Form of Prices at the Rate of Rs. X, Y And Z Respectively per Person. - Mathematics

Advertisements
Advertisements

प्रश्न

Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. xy and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of xy and z. What values are described in this equations?

उत्तर

\[A . T . Q\]
\[4x + 3y + 2z = 37000\]
\[5x + 3y + 4z = 47000\]
\[x + y + z = 12000\]
We can expressed these equations as AX = B .
\[\text{ Where }A = \begin{bmatrix}4 & 3 & 2 \\ 5 & 3 & 4 \\ 1 & 1 & 1\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}37000 \\ 47000 \\ 12000\end{bmatrix}\]
\[\left| A \right| = 4\left( 3 - 4 \right) - 3\left( 5 - 4 \right) + 2\left( 5 - 3 \right) = - 4 - 3 + 4 = - 3 \neq 0\]
So, A is non singular therefore inverse exists . 
\[ A_{11} = - 1 A_{12} = - 1 A_{13} = 2\]
\[ A_{21} = - 1 A_{22} = 2 A_{23} = - 1\]
\[ A_{31} = 6 A_{32} = - 6 A_{33} = - 3\]
\[adj A = \begin{bmatrix}- 1 & - 1 & 6 \\ - 1 & 2 & - 6 \\ 2 & - 1 & - 3\end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A = - \frac{1}{3}\begin{bmatrix}- 1 & - 1 & 6 \\ - 1 & 2 & - 6 \\ 2 & - 1 & - 3\end{bmatrix}\]
\[X = A^{- 1} B = - \frac{1}{3}\begin{bmatrix}- 1 & - 1 & 6 \\ - 1 & 2 & - 6 \\ 2 & - 1 & - 3\end{bmatrix} \begin{bmatrix}37000 \\ 47000 \\ 12000\end{bmatrix}\]
\[ = - \frac{1}{3}\begin{bmatrix}- 37000 - 47000 + 72000 \\ - 37000 + 94000 - 72000 \\ 74000 - 47000 - 36000\end{bmatrix} = - \frac{1}{3}\begin{bmatrix}- 12000 \\ - 15000 \\ - 9000\end{bmatrix} = \begin{bmatrix}4000 \\ 5000 \\ 3000\end{bmatrix}\]
\[So, x = 4000 , y = 5000\text{ and }z = 3000 .\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Solution of Simultaneous Linear Equations - Exercise 8.1 [पृष्ठ १७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 8 Solution of Simultaneous Linear Equations
Exercise 8.1 | Q 15 | पृष्ठ १७

संबंधित प्रश्‍न

Solve the system of linear equations using the matrix method.

x − y + z = 4

2x + y − 3z = 0

x + y + z = 2


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


Evaluate the following determinant:

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


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

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


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

\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}\]


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

\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]


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

\[\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y\end{vmatrix}\]


Evaluate :

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


\[\begin{vmatrix}b^2 + c^2 & ab & ac \\ ba & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2\end{vmatrix} = 4 a^2 b^2 c^2\]


Prove the following identities:
\[\begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix} = \left( 5x + \lambda \right) \left( \lambda - x \right)^2\]


Prove the following identity:

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

 


​Solve the following determinant equation:

\[\begin{vmatrix}1 & x & x^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{vmatrix} = 0, b \neq c\]

 


If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.


Find the value of \[\lambda\]  so that the points (1, −5), (−4, 5) and \[\lambda\]  are collinear.


If the points (x, −2), (5, 2), (8, 8) are collinear, find x using determinants.


Prove that :

\[\begin{vmatrix}z & x & y \\ z^2 & x^2 & y^2 \\ z^4 & x^4 & y^4\end{vmatrix} = \begin{vmatrix}x & y & z \\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4\end{vmatrix} = \begin{vmatrix}x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \\ x & y & z\end{vmatrix} = xyz \left( x - y \right) \left( y - z \right) \left( z - x \right) \left( x + y + z \right) .\]

 


Prove that :

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

\[\begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix} = \left( a^3 - 1 \right)^2\]

2y − 3z = 0
x + 3y = − 4
3x + 4y = 3


Solve each of the following system of homogeneous linear equations.
2x + 3y + 4z = 0
x + y + z = 0
2x − y + 3z = 0


Evaluate \[\begin{vmatrix}4785 & 4787 \\ 4789 & 4791\end{vmatrix}\]


Write the value of the determinant \[\begin{vmatrix}2 & - 3 & 5 \\ 4 & - 6 & 10 \\ 6 & - 9 & 15\end{vmatrix} .\]


If \[\begin{vmatrix}2x + 5 & 3 \\ 5x + 2 & 9\end{vmatrix} = 0\]


If |A| = 2, where A is 2 × 2 matrix, find |adj A|.


The value of the determinant

\[\begin{vmatrix}a^2 & a & 1 \\ \cos nx & \cos \left( n + 1 \right) x & \cos \left( n + 2 \right) x \\ \sin nx & \sin \left( n + 1 \right) x & \sin \left( n + 2 \right) x\end{vmatrix}\text{ is independent of}\]

 


Using the factor theorem it is found that a + bb + c and c + a are three factors of the determinant 

\[\begin{vmatrix}- 2a & a + b & a + c \\ b + a & - 2b & b + c \\ c + a & c + b & - 2c\end{vmatrix}\]
The other factor in the value of the determinant is


There are two values of a which makes the determinant  \[∆ = \begin{vmatrix}1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2a\end{vmatrix}\]  equal to 86. The sum of these two values is

 


Solve the following system of equations by matrix method:
3x + 4y − 5 = 0
x − y + 3 = 0


Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12


Solve the following system of equations by matrix method:
 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25


Solve the following system of equations by matrix method:
 2x + 6y = 2
3x − z = −8
2x − y + z = −3


Solve the following system of equations by matrix method:
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12


Show that each one of the following systems of linear equation is inconsistent:
2x + 5y = 7
6x + 15y = 13


If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ - 1 \\ 0\end{bmatrix}\], find x, y and z.

Solve the following system of equations by using inversion method

x + y = 1, y + z = `5/3`, z + x = `4/3`


Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.


Show that if the determinant ∆ = `|(3, -2, sin3theta),(-7, 8, cos2theta),(-11, 14, 2)|` = 0, then sinθ = 0 or `1/2`.


Let A = `[(1,sin α,1),(-sin α,1,sin α),(-1,-sin α,1)]`, where 0 ≤ α ≤ 2π, then:


The value (s) of m does the system of equations 3x + my = m and 2x – 5y = 20 has a solution satisfying the conditions x > 0, y > 0.


The system of simultaneous linear equations kx + 2y – z = 1,  (k – 1)y – 2z = 2 and (k + 2)z = 3 have a unique solution if k equals:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×