HomeOtherTS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(g)

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(g)

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I.
Examine whether the following systems of equations are consistent or inconsistent and if consistent find the complete solutions,
Question 1.
x + y + z = 4
2x + 5y – 2z = 3
x + 7y – 7z = 5

Answer:
Augmented matrix of the above system is
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g) 1
Rank of the matrix ρ(A) = 2 and ρ(AB) = 3.
Since ρ(A) ≠ ρ(AB), the given system of equa¬tions are inconsistent.

Question 2.
x + y + z = 6
x – y + z = 2
2x – y + 3z = 9

Answer:
Augmented matrix [AB] = 112111113629
Apply operations R2 – R1, R3 – 2R1, we get
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g) 2
Here ρ(A) = 3 and ρ(AB) = 3
Since ρ(A) = ρ(AB), the given system is consistent and has a unique solution.
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g) 3

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Question 3.
x + y + z = 1
2x + y + z = 2
x + 2y + 2z = 1 (March 2015-T.S)

Answer:
Augmented matrix of the system is
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g) 4
ρ(AB) = 2 and ρ(A) = 2 and ρ(A) = ρ(AB) < 3
The given system of equations is consis¬tent and has infinitely many solutions.
The given system is equivalent to x + y + z = 1 and y + z = 0.
Solution set is
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g) 5

Question 4.
x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0

Answer:
Augmented matrix of the system
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g) 6
Here ρ(A) = ρ(AB) = 3; and the system of given equations is consistent; and has a unique solution.
Also x = 1, y = 3, z = 5 form the solution.

Question 5.
x + y + z = 6
x + 2y + 3z = 10
x + 2y + 4z = 1

Answer:
Augmented matrix of the system is
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g) 7

Question 6.
x – 3y – 8z = – 10
3x + y – 4z = 0
2x + 5y + 6z = 13

Answer:
The augmented matrix of the above system is
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g) 8
Since ρ(A) = 2 = ρ(AB) < 3, given system of equations is consistent with infinitely many solutions.
The given system is equivalent to
x – 3y – 8z = – 10,
y + 2z = 3
Put z = t then y = 3 – 2t
∴ x = – 10 + 3(3 – 2t) + 8t
= -10 + 9 – 6t + 8t
= 2t – 1
Hence the solutions are given by
x = 2t – 1, y = 3 – 2t and z = t
Where t is any scalar.

Question 7.
2x + 3y + z = 9
x + 2y + 3z = 6
3x + y + 2z = 8

Answer:
Augmented matrix of the above system is
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g) 9
ρ(A) = ρ(AB) = 3; system is consistent and has a unique solution given by
x = 3518, y = 2918, z = 518

Question 8.
x + y + 4z = 6
3x + 2y – 2z = 9
5x + y + 2z = 13

Answer:
Augmented matrix of the system
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g) 10
ρ(A) = ρ(AB) = 3;
Hence the system is consistent and has a unique solution given by
x = 2, y = 2, z = 12.

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