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. x, y 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 x, y 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 .
\[ A_{11} = - 1 A_{12} = - 1 A_{13} = 2\]
\[ A_{21} = - 1 A_{22} = 2 A_{23} = - 1\]
\[ A_{31} = 6 A_{32} = - 6 A_{33} = - 3\]
\[adj A = \begin{bmatrix}- 1 & - 1 & 6 \\ - 1 & 2 & - 6 \\ 2 & - 1 & - 3\end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A = - \frac{1}{3}\begin{bmatrix}- 1 & - 1 & 6 \\ - 1 & 2 & - 6 \\ 2 & - 1 & - 3\end{bmatrix}\]
\[X = A^{- 1} B = - \frac{1}{3}\begin{bmatrix}- 1 & - 1 & 6 \\ - 1 & 2 & - 6 \\ 2 & - 1 & - 3\end{bmatrix} \begin{bmatrix}37000 \\ 47000 \\ 12000\end{bmatrix}\]
\[ = - \frac{1}{3}\begin{bmatrix}- 37000 - 47000 + 72000 \\ - 37000 + 94000 - 72000 \\ 74000 - 47000 - 36000\end{bmatrix} = - \frac{1}{3}\begin{bmatrix}- 12000 \\ - 15000 \\ - 9000\end{bmatrix} = \begin{bmatrix}4000 \\ 5000 \\ 3000\end{bmatrix}\]
\[So, x = 4000 , y = 5000\text{ and }z = 3000 .\]
APPEARS IN
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:
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 :
Prove that :
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
Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant
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
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: