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.

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?

Advertisements

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 7: Solution of Simultaneous Linear Equations - Exercise 8.1 [पृष्ठ १७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 7 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 + 2z = 7

3x + 4y − 5z = −5

2x − y + 3z = 12


Find the value of x, if

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


Evaluate the following determinant:

\[\begin{vmatrix}1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9\end{vmatrix}\]


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:

\[\begin{vmatrix}1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2\end{vmatrix}\]


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

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


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

\[\begin{vmatrix}\sqrt{23} + \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{46} & 5 & \sqrt{10} \\ 3 + \sqrt{115} & \sqrt{15} & 5\end{vmatrix}\]


\[If ∆ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix}, ∆_1 = \begin{vmatrix}1 & 1 & 1 \\ yz & zx & xy \\ x & y & z\end{vmatrix},\text{ then prove that }∆ + ∆_1 = 0 .\]


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


Without expanding, prove that

\[\begin{vmatrix}a & b & c \\ x & y & z \\ p & q & r\end{vmatrix} = \begin{vmatrix}x & y & z \\ p & q & r \\ a & b & c\end{vmatrix} = \begin{vmatrix}y & b & q \\ x & a & p \\ z & c & r\end{vmatrix}\]


\[If \begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix} = 0,\text{ find the value of }\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}, p \neq a, q \neq b, r \neq c .\]

 


​Solve the following determinant equation:

\[\begin{vmatrix}x + a & x & x \\ x & x + a & x \\ x & x & x + a\end{vmatrix} = 0, a \neq 0\]

 


Show that
`|(x-3,x-4,x-alpha),(x-2,x-3,x-beta),(x-1,x-2,x-gamma)|=0`, where α, β, γ are in A.P.

 


Using determinants prove that the points (ab), (a', b') and (a − a', b − b') are collinear if ab' = a'b.

 

Prove that :

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

 


Prove that :

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

Given: x + 2y = 1
            3x + y = 4


6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8


2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11


x + y + z + 1 = 0
ax + by + cz + d = 0
a2x + b2y + x2z + d2 = 0


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


Solve each of the following system of homogeneous linear equations.
3x + y + z = 0
x − 4y + 3z = 0
2x + 5y − 2z = 0


Find the value of the determinant
\[\begin{bmatrix}4200 & 4201 \\ 4205 & 4203\end{bmatrix}\]


If \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 0 \\ - 1 & 0\end{bmatrix}\] , find |AB|.

 

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


If I3 denotes identity matrix of order 3 × 3, write the value of its determinant.


Find the value of x from the following : \[\begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix} = 0\]


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


If xyare different from zero and \[\begin{vmatrix}1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z\end{vmatrix} = 0\] , then the value of x−1 + y−1 + z−1 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 + 4y + 7z = 14

2x − y + 3z = 4

x + 2y − 3z = 0


Show that the following systems of linear equations is consistent and also find their solutions:
x − y + z = 3
2x + y − z = 2
−x −2y + 2z = 1


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

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


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


The value of x, y, z for the following system of equations x + y + z = 6, x − y+ 2z = 5, 2x + y − z = 1 are ______


`abs ((2"xy", "x"^2, "y"^2),("x"^2, "y"^2, 2"xy"),("y"^2, 2"xy", "x"^2)) =` ____________.


In system of equations, if inverse of matrix of coefficients A is multiplied by right side constant B vector then resultant will be?


What is the nature of the given system of equations

`{:(x + 2y = 2),(2x + 3y = 3):}`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×