Question 12:

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. C purchases 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.

Answer:

$The prices of three commodities P, Q and R are Rs x, Rs y and Rs z per unit, respectively .$

$According to the question, 3 x + 5 y - 4 z = 6000 2 x - 3 y + z = 5000 - x + 4 y + 6 z = 13000 The given system of equations can be written in matrix form as follows: [\begin{matrix} 3 & 5 & - 4 \\ 2 & - 3 & 1 \\ - 1 & 4 & 6 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 6000 \\ 5000 \\ 13000 \end{matrix}] A X = B Here, A = [\begin{matrix} 3 & 5 & - 4 \\ 2 & - 3 & 1 \\ - 1 & 4 & 6 \end{matrix}] X = [\begin{matrix} x \\ y \\ z \end{matrix}] B = [\begin{matrix} 6000 \\ 5000 \\ 13000 \end{matrix}] Now, |A| = 3 (- 18 - 4) - 5 (12 + 1) - 4 (8 - 3) = - 66 - 65 - 20 = - 151 \neq 0 So, A^{- 1} exists . {Let C}_{i j} {be the cofactors of the elements a}_{i j} in A= [a_{i j}] . Then, C_{11} = {(- 1)}^{1 + 1} |\begin{matrix} - 3 & 1 \\ 4 & 6 \end{matrix}| = - 22, C_{12} = {(- 1)}^{1 + 2} |\begin{matrix} 2 & 1 \\ - 1 & 6 \end{matrix}| = - 13, C_{13} = {(- 1)}^{1 + 3} |\begin{matrix} 2 & - 3 \\ - 1 & 4 \end{matrix}| = 5 C_{21} = {(- 1)}^{2 + 1} |\begin{matrix} 5 & - 4 \\ 4 & 6 \end{matrix}| = - 46, C_{22} = {(- 1)}^{2 + 2} |\begin{matrix} 3 & - 4 \\ - 1 & 6 \end{matrix}| = 14, C_{23} = {(- 1)}^{2 + 3} |\begin{matrix} 3 & 5 \\ - 1 & 4 \end{matrix}| = - 17 C_{31} = {(- 1)}^{3 + 1} |\begin{matrix} 5 & - 4 \\ - 3 & 1 \end{matrix}| = - 7, C_{32} = {(- 1)}^{3 + 2} |\begin{matrix} 3 & - 4 \\ 2 & 1 \end{matrix}| = - 11, C_{33} = {(- 1)}^{3 + 3} |\begin{matrix} 3 & 5 \\ 2 & - 3 \end{matrix}| = - 19 adj A = {[\begin{matrix} - 22 & - 13 & 5 \\ - 46 & 14 & - 17 \\ - 7 & - 11 & - 19 \end{matrix}]}^{T} = [\begin{matrix} - 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19 \end{matrix}] \Rightarrow A^{- 1} = \frac{1}{|A|} adj A = \frac{1}{- 151} [\begin{matrix} - 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19 \end{matrix}] X = A^{- 1} B \Rightarrow X = \frac{1}{- 151} [\begin{matrix} - 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19 \end{matrix}] [\begin{matrix} 6000 \\ 5000 \\ 13000 \end{matrix}] \Rightarrow X = \frac{1}{- 151} [\begin{matrix} - 453000 \\ - 151000 \\ - 302000 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 3000 \\ 1000 \\ 2000 \end{matrix}] ∴ x = 3000, y = 1000 and z = 2000$
$Thus, the prices of the three commodities P, Q and R are Rs 3000, Rs 1000 and Rs 2000 per unit, respectively .$

Page No 7.17:

Question 13:

The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.

Answer:

As per the information given in the question, the following equations hold true:

The above three equations can be represented in the form of a matrix as

Or AX = B, where,

Thus, A is non-singular. Therefore, its inverse exists.

Adj A is given by∴

Therefore, the number of awardees for Honesty, Cooperation and Supervision are 3, 4, and 5 respectively.

One more value which the management of the colony must include for awards may be Sincerity.

Page No 7.17:

Question 14:

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.

Answer:

Let the award money given for Honesty, Regularity and Hard work be x, y and z respectively.

Since total cash award is Rs 6,000.

∴ x + y + z = Rs 6,000 ...(1)

Three times the award money for Hard work and Honesty is Rs 11,000.

∴ x + 3 z = Rs 11,000

⇒ x + 0.y + 3 z = Rs 11,000 ...(2)

Award money for Honesty and Hard work is double the one given for regularity.

∴ x + z = 2y

⇒ x − 2y + z = 0 ...(3)

The above system can be written in matrix form as,

Or AX = B, where

Thus, A is non-singular. Hence, it is invertible.

Adj A =

If $A = [\begin{matrix} 2 & 4 \\ 4 & 3 \end{matrix}], X = [\begin{matrix} n \\ 1 \end{matrix}], B = [\begin{matrix} 8 \\ 11 \end{matrix}]$ and AX = B, then find n.

Answer:

$Here, [\begin{matrix} 2 & 4 \\ 4 & 3 \end{matrix}] [\begin{matrix} n \\ 1 \end{matrix}] = [\begin{matrix} 8 \\ 11 \end{matrix}] \Rightarrow [\begin{matrix} 2 n + 4 \\ 4 n + 3 \end{matrix}] = [\begin{matrix} 8 \\ 11 \end{matrix}] \Rightarrow 2 n + 4 = 8 \Rightarrow 2 n = 4 \Rightarrow n = 2$

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