Advertisements
Advertisements
Question
Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A 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 ₹1,600. School B wants to spend ₹2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ₹900, 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 sincerity, truthfulness and helpfulness be ₹x, ₹y and ₹z respectively.
Since, the total cash award is ₹900.
∴ x + y + z = 900 ....(1)
Award money given by school A is ₹1,600.
∴ 3x + 2y + z = 1600 ....(2)
Award money given by school B is ₹2,300.
∴ 4x + y + 3z = 2300 ....(3)
The above system of equations can be written in matrix form CX = D as
\[\begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}900 \\ 1600 \\ 2300\end{bmatrix}\]
\[\text{ Where,} C = \begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }D = \begin{bmatrix}900 \\ 1600 \\ 2300\end{bmatrix}\]
Now,
\[\left| C \right| = \begin{vmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{vmatrix}\]
\[ = 1\left( 6 - 1 \right) - 1\left( 9 - 4 \right) + 1(3 - 8)\]
\[ = 5 - 5 - 5\]
\[ = - 5\]
\[\text{ Let }C_{ij}\text{ be the cofactors of elements }c_{ij}\text{ in }C = \left[ c_{ij} \right] . \text{ Then, }\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 1 \\ 1 & 3\end{vmatrix} = 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & 1 \\ 4 & 3\end{vmatrix} = - 5, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix} = - 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 3\end{vmatrix} = - 2 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 4 & 3\end{vmatrix} = - 1 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 4 & 1\end{vmatrix} = 3\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} = - 1\]
\[adj C = \begin{bmatrix}5 & - 5 & - 5 \\ - 2 & - 1 & 3 \\ - 1 & 2 & - 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\]
\[ \Rightarrow C^{- 1} = \frac{1}{\left| C \right|}adj C\]
\[ = \frac{1}{- 5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\]
\[X = C^{- 1} D\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\begin{bmatrix}900 \\ 1600 \\ 2300\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}4500 - 3200 - 2300 \\ - 4500 - 1600 + 4600 \\ - 4500 + 4800 - 2300\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}- 1000 \\ - 1500 \\ - 2000\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 1000}{- 5}, y = \frac{- 1500}{- 5}\text{ and }z = \frac{- 2000}{- 5}\]
\[ \therefore x = 200, y = 300\text{ and }z = 400 .\]
Hence, the award money for each value of sincerity, truthfulness and helpfulness is ₹200, ₹300 and ₹400.
One more value which should be considered for award hardwork.
APPEARS IN
RELATED QUESTIONS
Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
Solve the system of linear equations using the matrix method.
5x + 2y = 3
3x + 2y = 5
Solve the system of linear equations using the matrix method.
2x + y + z = 1
x – 2y – z = `3/2`
3y – 5z = 9
Solve the system of the following equations:
`2/x+3/y+10/z = 4`
`4/x-6/y + 5/z = 1`
`6/x + 9/y - 20/x = 2`
Evaluate the following determinant:
\[\begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix}\]
Evaluate
\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]
If \[A = \begin{bmatrix}2 & 5 \\ 2 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}4 & - 3 \\ 2 & 5\end{bmatrix}\] , verify that |AB| = |A| |B|.
For what value of x the matrix A is singular?
\[A = \begin{bmatrix}x - 1 & 1 & 1 \\ 1 & x - 1 & 1 \\ 1 & 1 & x - 1\end{bmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25\end{vmatrix}\]
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:
\[\begin{vmatrix}a + b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b\end{vmatrix}\]
Evaluate the following:
\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix}\]
\[If ∆ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix}, ∆_1 = \begin{vmatrix}1 & 1 & 1 \\ yz & zx & xy \\ x & y & z\end{vmatrix},\text{ then prove that }∆ + ∆_1 = 0 .\]
Find the area of the triangle with vertice at the point:
(3, 8), (−4, 2) and (5, −1)
Find values of k, if area of triangle is 4 square units whose vertices are
(−2, 0), (0, 4), (0, k)
Prove that :
Prove that :
Prove that :
3x + y = 19
3x − y = 23
5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7
Find the value of the determinant
\[\begin{bmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{bmatrix}\]
If \[A = \begin{bmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{bmatrix}\]. Write the cofactor of the element a32.
If \[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}\] , write the value of x.
If \[\begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8\] , write the value of x.
If \[A + B + C = \pi\], then the value of \[\begin{vmatrix}\sin \left( A + B + C \right) & \sin \left( A + C \right) & \cos C \\ - \sin B & 0 & \tan A \\ \cos \left( A + B \right) & \tan \left( B + C \right) & 0\end{vmatrix}\] is equal to
The number of distinct real roots of \[\begin{vmatrix}cosec x & \sec x & \sec x \\ \sec x & cosec x & \sec x \\ \sec x & \sec x & cosec x\end{vmatrix} = 0\] lies in the interval
\[- \frac{\pi}{4} \leq x \leq \frac{\pi}{4}\]
Let \[A = \begin{bmatrix}1 & \sin \theta & 1 \\ - \sin \theta & 1 & \sin \theta \\ - 1 & - \sin \theta & 1\end{bmatrix},\text{ where 0 }\leq \theta \leq 2\pi . \text{ Then,}\]
Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12
Solve the following system of equations by matrix method:
x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1
Show that each one of the following systems of linear equation is inconsistent:
4x − 2y = 3
6x − 3y = 5
Show that each one of the following systems of linear equation is inconsistent:
x + y − 2z = 5
x − 2y + z = −2
−2x + y + z = 4
The prices of three commodities P, Q and R are Rs x, y and z per unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchases 3 units of Q and sells 2 units of P and 1 unit of R. Cpurchases 1 unit of P and sells 4 units of Q and 6 units of R. In the process A, B and C earn Rs 6000, Rs 5000 and Rs 13000 respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using matrix method.
A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.
2x + 3y − z = 0
x − y − 2z = 0
3x + y + 3z = 0
The system of equation x + y + z = 2, 3x − y + 2z = 6 and 3x + y + z = −18 has
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
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 the system of equations x + ky - z = 0, 3x - ky - z = 0 & x - 3y + z = 0 has non-zero solution, then k is equal to ____________.
Let A = `[(1,sin α,1),(-sin α,1,sin α),(-1,-sin α,1)]`, where 0 ≤ α ≤ 2π, then:
The existence of unique solution of the system of linear equations x + y + z = a, 5x – y + bz = 10, 2x + 3y – z = 6 depends on
