English

Two Schools P And Q Want to Award Their Selected Students on the Values of Tolerance, Kindness and Leadership - Mathematics

Advertisements
Advertisements

Question

Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award ₹x each, ₹y each and ₹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹2,200. School Q wants to spend ₹3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is ₹1,200, using matrices, find the award money for each value.
Apart from these three values, suggest one more value which should be considered for award.

Advertisements

Solution

​​Let the award money given for Tolerance, Kindness and Leadership be ₹x, ₹y and ₹z respectively.

Since, the total cash award is ₹1,200.
∴ x + y + z = 1200                    ....(1)

Award money given by school P is ₹2,200.
∴ 3x + 2y + z = 2200                ....(2)

Award money given by school Q is ₹3,100.
∴ 4x + y + 3z = 3100                ....(3)

The above system of equations can be written in matrix form AX = B as

\[\begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1200 \\ 2200 \\ 3100\end{bmatrix}\]
\[\text{ Where, }A = \begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}1200 \\ 2200 \\ 3100\end{bmatrix}\]
Now, 
\[\left| A \right| = \begin{vmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{vmatrix}\]
\[ = 1\left( 6 - 1 \right) - 1\left( 9 - 4 \right) + 1(3 - 8)\]
\[ = 5 - 5 - 5\]
\[ = - 5\]
\[\text{ Let }C_{ij}\text{ be the cofactors of elements }a_{ij}\text{ in }A = \left[ a_{ij} \right] .\text{ Then,} \]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 1 \\ 1 & 3\end{vmatrix} = 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & 1 \\ 4 & 3\end{vmatrix} = - 5, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix} = - 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 3\end{vmatrix} = - 2 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 4 & 3\end{vmatrix} = - 1 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 4 & 1\end{vmatrix} = 3\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} = - 1\]
\[adj A = \begin{bmatrix}5 & - 5 & - 5 \\ - 2 & - 1 & 3 \\ - 1 & 2 & - 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\begin{bmatrix}1200 \\ 2200 \\ 3100\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}6000 - 4400 - 3100 \\ - 6000 - 2200 + 6200 \\ - 6000 + 6600 - 3100\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}- 1500 \\ - 2000 \\ - 2500\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 1500}{- 5}, y = \frac{- 2000}{- 5}\text{ and }z = \frac{- 2500}{- 5}\]
\[ \therefore x = 300, y = 400\text{ and }z = 500 .\]

Hence, the award money for each value of Tolerance, Kindness and Leadership is ₹300, ₹400 and ₹500.
​One more value which should be considered for award is Honesty.

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Solution of Simultaneous Linear Equations - Exercise 8.1 [Page 17]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 8 Solution of Simultaneous Linear Equations
Exercise 8.1 | Q 19 | Page 17

RELATED QUESTIONS

If `|[2x,5],[8,x]|=|[6,-2],[7,3]|`, write the value of x.


Find the value of a if `[[a-b,2a+c],[2a-b,3c+d]]=[[-1,5],[0,13]]`


If `|[x+1,x-1],[x-3,x+2]|=|[4,-1],[1,3]|`, then write the value of x.


Evaluate the following determinant:

\[\begin{vmatrix}x & - 7 \\ x & 5x + 1\end{vmatrix}\]


Evaluate the following determinant:

\[\begin{vmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{vmatrix}\]


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


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

\[\begin{vmatrix}a + b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b\end{vmatrix}\]


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

\[\begin{vmatrix}\sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1\end{vmatrix}, where A, B, C \text{ are the angles of }∆ ABC .\]


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


​Solve the following determinant equation:

\[\begin{vmatrix}3x - 8 & 3 & 3 \\ 3 & 3x - 8 & 3 \\ 3 & 3 & 3x - 8\end{vmatrix} = 0\]

 


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

xy = 5
y + z = 3
x + z = 4


2y − 3z = 0
x + 3y = − 4
3x + 4y = 3


5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7


x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0


A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission 

Month Sale of units Total commission
drawn (in Rs)
  A B C  
Jan 90 100 20 800
Feb 130 50 40 900
March 60 100 30 850


Find out the rates of commission on items A, B and C by using determinant method.


For what value of x, the following matrix is singular?

\[\begin{bmatrix}5 - x & x + 1 \\ 2 & 4\end{bmatrix}\]

 


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


Find the value of the determinant 
\[\begin{bmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{bmatrix}\]

 


If \[A = \begin{bmatrix}0 & i \\ i & 1\end{bmatrix}\text{  and }B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\] , find the value of |A| + |B|.


If A = [aij] is a 3 × 3 scalar matrix such that a11 = 2, then write the value of |A|.

 

Solve the following system of equations by matrix method:
 x − y + z = 2
2x − y = 0
2y − z = 1


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


Show that the following systems of linear equations is consistent and also find their solutions:
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3


x + y − 6z = 0
x − y + 2z = 0
−3x + y + 2z = 0


If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}2 \\ - 1 \\ 3\end{bmatrix}\], find x, y, z.

Consider the system of equations:
a1x + b1y + c1z = 0
a2x + b2y + c2z = 0
a3x + b3y + c3z = 0,
if \[\begin{vmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix}\]= 0, then the system has


Let a, b, c be positive real numbers. The following system of equations in x, y and z 

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1, \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1, - \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \text { has }\]
(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions

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

x +y + z = 6

y + 3z = 11

and x -2y +z = 0


System of equations x + y = 2, 2x + 2y = 3 has ______


Three chairs and two tables cost ₹ 1850. Five chairs and three tables cost ₹2850. Find the cost of four chairs and one table by using matrices


Solve the following equations by using inversion method.

x + y + z = −1, x − y + z = 2 and x + y − z = 3


The cost of 4 dozen pencils, 3 dozen pens and 2 dozen erasers is ₹ 60. The cost of 2 dozen pencils, 4 dozen pens and 6 dozen erasers is ₹ 90. Whereas the cost of 6 dozen pencils, 2 dozen pens and 3 dozen erasers is ₹ 70. Find the cost of each item per dozen by using matrices


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


If `alpha, beta, gamma` are in A.P., then `abs (("x" - 3, "x" - 4, "x" - alpha),("x" - 2, "x" - 3, "x" - beta),("x" - 1, "x" - 2, "x" - gamma)) =` ____________.


If `|(x + 1, x + 2, x + a),(x + 2, x + 3, x + b),(x + 3, x + 4, x + c)|` = 0, then a, b, care in


If the system of linear equations

2x + y – z = 7

x – 3y + 2z = 1

x + 4y + δz = k, where δ, k ∈ R has infinitely many solutions, then δ + k is equal to ______.


Let P = `[(-30, 20, 56),(90, 140, 112),(120, 60, 14)]` and A = `[(2, 7, ω^2),(-1, -ω, 1),(0, -ω, -ω + 1)]` where ω = `(-1 + isqrt(3))/2`, and I3 be the identity matrix of order 3. If the determinant of the matrix (P–1AP – I3)2 is αω2, then the value of α is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×