# RD SHARMA Solutions for Class 12-science Maths Chapter 8 - Solution of Simultaneous Linear Equations

## Chapter 8 - Solution of Simultaneous Linear Equations Exercise Ex. 8.1

Using A^{-1}, solve the system of linear equations

X - 2y = 10, 2x - y - z = 8 and -2y + z = 7

The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping and 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 must include for awards.

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs. 6000. Three times the award money for Hard work added to that given for honesty amounts to Rs. 11000. 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 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.

The school can include an award for creativity and extra-curricular activities.

Two institutions decided to award
their employees for the three values of resourcefulness, competence and
determination in the form of prizes at the rate of Rs. *x*, Rs. *y* and Rs. *z* respectively per person. The first
institution decided to award respectively 4, 3 and 2 employees with a total prize money of Rs. 37000 and the second
institution decided to award respectively 5, 3 and 4 employees with a total
prize money of Rs. 47000. If all the three prizes per person together amount
to Rs. 12000, then using matrix method find the value of *x*, *y* and *z*. What values are described in these
equations?

Two factories decided
to award their employees for three values of (a) adaptable to new techniques,
(b) careful and alert in difficult situations and (c) keeping calm in tense
situations, at the rate of Rs. *x*,
Rs. *y* and Rs. *z* per person respectively. The first factory decided to honour respectively 2, 4 and 3 employees with a total prize money of Rs. 29000. The second factory
decided to honour respectively 5, 2 and 3 employees
with the prize money of Rs. 30500. If the three prizes per person together
cost Rs 9500, then

(i) represent the above situation by matrix equation and form linear equations using matrix multiplication.

(ii) Solve these equations using matrices.

(iii) Which values are reflected in the questions?

Keeping calm in a tense situation is more rewarding than carefulness, and carefulness is more rewarding than adaptability.

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 Rs. *x* each Rs. *y* each and
Rs. *z* each for the three respective
values to 3, 2 and 1 students respectively with a total
award money of Rs. 1,600. School B wants to spend Rs 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 Rs 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.

Two schools *P* and *Q* want to award
their selected students on the values of Discipline, Politeness and
Punctuality. The school *P* wants to
award Rs. *x* each, Rs. *y* each and Rs. *z* each for the three respectively values to its 3, 2 and 1
students with a total award money of Rs. 1,000. School *Q* wants to spend Rs. 1,500 to award its 4, 1 and 3 students on
the respective values (by giving the same award money for three values as
before). If the total amount of awards for one prize on each value is Rs.
600, using matrices, find the award money for each value. Apart from the
above three values, suggest one more value for awards.

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 Rs. *x* each, Rs. *y* each and Rs. *z* each
for the three respectively values to its 3, 2 and 1 students with a total
award money of Rs. 2,200. School *Q*
wants to spend Rs. 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 Rs. 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.

A total amount of Rs. 7000 is deposited in three different saving bank accounts with annual interest rates 5%, 8% and 8.5% respectively. The total annual interest from these three accounts is Rs. 550. Equal amounts have been deposited in the 5% and 8% savings accounts. Find the amount deposited in each of the three accounts, with the help of matrices.

Let the amount deposited be x, y and z respectively.

As per the data in the question, we get

If
find A^{-1}
and hence solve the system of equations 2x + y - 3z = 13, 3x + 2y + z = 4, x
+ 2y - z = 8.

Therefore, A is invertible.

Let C_{ij} be the co-factors of the elements a_{ij}.

Now, the given system of equations is expressible as

Here we have |A^{T}|
= |A| = -16 ≠ 0

Therefore, the given system of equations is consistent with a unique solution given by

Use the product to solve the system of equations x + 3z = -9, -x + 2y - 2z = 4, 2x - 3y + 4z = -3.

Let

Now,

Now, the given system of equations is expressible as

Here we have |B^{T}|
= |B| = -1 ≠ 0

Therefore, the given system of equations is consistent with a unique solution given by

Hence, x = 36, y = 5 and z = -15.

A shopkeeper has 3 varieties of pens 'A', 'B' and 'C'. Meenu purchased 1 pen of each variety for a total of Rs. 21. Jeen purchased 4 pens of 'A' variety, 3 pens of 'B' variety and 2 pens of 'C' variety for Rs. 60. Using matrix method find the cost of each pen.

From the given information, we can form a matrix as follows

Applying R_{2}→ R_{2}
- 4R_{1}, R_{3}→ R_{3} - 6R_{1}

Applying R_{3}→ R_{3}
+ (-4R_{1})

From the above matrix form, we get

A + B + C = 21 … (i)

-B - 2C = -24 … (ii)

5C = 40

⇒ C = 8 … (iii)

Putting the value of C in (ii), we get

B = 8

Substituting B and C in (i), we get

C = 5

Hence, cost of variety 'A' pen is Rs. 8, cost of variety B pen is Rs. 8 and cost of variety 'C' pen is Rs. 5.

## Chapter 8 - Solution of Simultaneous Linear Equations Exercise Ex. 8.2

## Chapter 8 - Solution of Simultaneous Linear Equations Exercise MCQ

The system of equation x + y + z = 2, 3x - y + 2z = 6 and 3x + y + z = -18 has

a. a unique solution

b. no solution

c. an infinite number of solutions

d. zero solution as the only solution

a. 3

b. 2

c. 1

d. 0

a.

b.

c.

d.

The number of solutions of the system of equations:

, is

a. 3

b. 2

c. 1

d. 0

The system of linear equations:

x + y + z = 2

2x + y - z = 3

3x + 2y + kz = 4

Has a unique solution if

a. k ≠ 0

b. -1 < k < 1

c. -2 < k < 2

d. k = 0

Consider the system of equations:

a_{1}x + b_{1}y + c_{1}z = 0

a_{2}x + b_{2}y + c_{2}z = 0

a_{3}x + b_{3}y + c_{3}z = 0.

a. more than two solutions

b. one trivial and one non-trivial solutions

c. no solution

d. only trivial solution (0, 0, 0)

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

a. no solutions

b. unique solution

c. infinitely many solutions

d. finitely many solutions

For the system of equations :

x + 2y + 3z = 1

2x + y + 3z = 2

5x + 5y + 9z = 4

a. there is only one solution

b. there exists infinitely many solution

c. there is no solution

d. none of these

Correct optio : (a)

x + 2y + 3z = 1

2x + y + 3z = 2

5x + 5y + 9z = 4

The determinant of the coefficient matrix is

= -6-2 (18 - 15) + 3(10 - 5)

= -6 - 6 + 15

= 3 ≠ 0

The right hand side is also non zero.

The system has a unique solution.

The existence of the unique solution of the system of equations:

x + y +z = λ

5x - y + μz = 10

2x + 3y - z = 6 depends on

a. μ only

b. λ only

c. λ and μ both

d. neither λ nor μ

The system of equations:

x + y + z = 5

x + 2y + 3z = 9

x + 3y + λz = μ

Has a unique solution, if

a. λ = 5, μ = 13

b. λ ≠ 5

c. λ = 5, μ ≠ 13

d. μ ≠ 13

## Chapter 8 - Solution of Simultaneous Linear Equations Exercise Ex. 8VSAQ

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