Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(h) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(h)

Solve the following systems of equations.
(i) by using Cramer’s rule and matrix inversion method, when the coefficient matrix is non-singular.
(ii) by using the Gauss-Jordan method. Also, determine whether the system has a unique solution or an infinite number of solutions, or no solution, and find the solutions if exist.

Question 1.
5x – 6y + 4z = 15
7x + 4y – 3z = 19
2x + y + 6z = 46
Solution:
(i) Cramer’s rule:
Δ = $$\left|\begin{array}{ccc} 5 & -6 & 4 \\ 7 & 4 & -3 \\ 2 & 1 & 6 \end{array}\right|$$
= 5(24 + 3) + 6(42 + 6) + 4(7 – 8)
= 135 + 288 – 4
= 419
Δ1 = $$\left|\begin{array}{ccc} 15 & -6 & 4 \\ 19 & 4 & -3 \\ 46 & 1 & 6 \end{array}\right|$$
= 15(24 + 3) + 6(114 + 138) + 4(19 – 184)
= 405 + 1512 – 660
= 1917 – 660
= 1257
Δ2 = $$\left|\begin{array}{ccc} 5 & 15 & 4 \\ 7 & 19 & -3 \\ 2 & 46 & 6 \end{array}\right|$$
= 5(114 + 138) – 15(42 + 6) + 4(322 – 38)
= 1260 – 720 + 1136
= 1676
Δ3 = $$\left|\begin{array}{ccc} 5 & -6 & 15 \\ 7 & 4 & 19 \\ 2 & 1 & 46 \end{array}\right|$$
= 5(184 – 19) + 6(322 – 38) + 15(7 – 8)
= 825 + 1704 – 15
= 2529 – 15
= 2514 Solution is x = 3, y = 4, z = 6.

(ii) Matrix inversion method:   Solution is x = 3, y = 4, z = 6

(iii) Gauss-Jordan method: ∴ Unique solution exists.
Solution is x = 3, y = 4, z = 6. Question 2.
x + y + z = 1
2x + 2y + 3z = 6
x + 4y + 9z = 3
Solution:
(i) Cramer’s rule:
Δ = $$\left|\begin{array}{lll} 1 & 1 & 1 \\ 2 & 2 & 3 \\ 1 & 4 & 9 \end{array}\right|$$
= 1(18 – 12) – 1(18 – 3) + 1(8 – 2)
= 6 – 15 + 6
= -3
Δ1 = $$\left|\begin{array}{lll} 1 & 1 & 1 \\ 6 & 2 & 3 \\ 3 & 4 & 9 \end{array}\right|$$
= 1(18 – 12) – 1(54 – 9) + 1(24 – 6)
= 6 – 45 + 18
= -21
Δ2 = $$\left|\begin{array}{lll} 1 & 1 & 1 \\ 2 & 6 & 3 \\ 1 & 3 & 9 \end{array}\right|$$
= 1(54 – 9) – 1(18 – 3) + 1(6 – 6)
= 45 – 15
= 30
Δ3 = $$\left|\begin{array}{lll} 1 & 1 & 1 \\ 2 & 2 & 6 \\ 1 & 4 & 3 \end{array}\right|$$
= 1(6 – 24) – 1(6 – 6) + 1(8 – 2)
= -18 – 0 + 6
= -12 Solution is x = 7, y = -10, z = 4

(ii) Matrix inversion method:   ∴ Solution is x = 7, y = -10, z = 4

(iii) Gauss-Jordan method:
Augmented matrix is A = $$\left[\begin{array}{llll} 1 & 1 & 1 & 1 \\ 2 & 2 & 3 & 6 \\ 1 & 4 & 9 & 3 \end{array}\right]$$
R2 → R2 – 2R1, R3 → R3 – R1 Unique solution exists.
∴ Solution is x = 7, y = -10, z = 4 Question 3.
x – y + 3z = 5
4x + 2y – z = 0
-x + 3y + z = 5
Solution:
(i) Cramer’s rule:
Δ = $$\left|\begin{array}{ccc} 1 & -1 & 3 \\ 4 & 2 & -1 \\ -1 & 3 & 1 \end{array}\right|$$
= 1(2 + 3) + 1(4 – 1) + 3(12 + 2)
= 5 + 3 + 42
= 50
Δ1 = $$\left|\begin{array}{ccc} 5 & -1 & 3 \\ 0 & 2 & -1 \\ 5 & 3 & 1 \end{array}\right|$$
= 5(2 + 3) + 1(0 + 5) + 3(0 – 10)
= 25 + 5 – 30
= 0
Δ2 = $$\left|\begin{array}{ccc} 1 & 5 & 3 \\ 4 & 0 & -1 \\ -1 & 5 & 1 \end{array}\right|$$
= 1(0 + 5) – 5(4 – 1) + 3(20 – 0)
= 5 – 15 + 60
= 50
Δ3 = $$\left|\begin{array}{ccc} 1 & -1 & 5 \\ 4 & 2 & 0 \\ -1 & 3 & 5 \end{array}\right|$$
= 1(10 – 0) + 1(20 – 0) + 5(12 + 2)
= 10 + 20 + 70
= 100 ∴ Solution is x = 0, y = 1, z = 2.

(ii) Matrix inversion method:   Solution is x = 0, y = 1, z = 2

(iii) Gauss Jordan method: Unique solution exists.
∴ Solution is x = 0, y = 1, z = 2 Question 4.
2x + 6y + 11 = 0
6x + 20y – 6z + 3 = 0
6y – 18z + 1 = 0
Solution:
(i) Cramer’s rule:
Δ = $$\left|\begin{array}{ccc} 2 & 6 & 0 \\ 6 & 20 & -6 \\ 0 & 6 & -18 \end{array}\right|$$
= 2(-360 + 36) – 6(-108 – 0)
= -648 + 648
= 0
∴ Cramer’s rule and matrix inversion method cannot be used.
∵ Δ = 0

(ii) Gauss Jordan method: ρ(A) = 2, ρ(AB) = 3
ρ(A) ≠ ρ(AB)
∴ The given system of equations does not have a solution. Question 5.
2x – y + 3z = 9
x + y + z = 6
x – y + z = 2
Solution:
(i) Cramer’s rule:
Δ = $$\left|\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{array}\right|$$
= 2(1 + 1) + 1(1 – 1) + 3(-1 – 1)
= 4 + 0 – 6
= -2
Δ1 = $$\left|\begin{array}{ccc} 9 & -1 & 3 \\ 6 & 1 & 1 \\ 2 & -1 & 1 \end{array}\right|$$
= 9(1 + 1) + 1(6 – 2) + 3(-6 – 2)
= 18 + 4 – 24
= -2
Δ2 = $$\left|\begin{array}{lll} 2 & 9 & 3 \\ 1 & 6 & 1 \\ 1 & 2 & 1 \end{array}\right|$$
= 2(6 – 2) – 9(1 – 1) + 3(2 – 6)
= 8 – 0 – 12
= -4
Δ3 = $$\left|\begin{array}{ccc} 2 & -1 & 9 \\ 1 & 1 & 6 \\ 1 & -1 & 2 \end{array}\right|$$
= 2(2 + 6) + 1(2 – 6) + 9(-1 – 1)
= 16 – 4 – 18
= -6 Solution is x = 1, y = 2, z = 3.

(ii) Matrix inversion method:  Solution is x = 1, y = 2, z = 3.

(iii) Gauss-Jordan method: ∴ The given equations have a unique solution.
Solution is x = 1, y = 2, z = 3 Question 6.
2x – y + 8z = 13
3x + 4y + 5z = 18
5x – 2y + 7z = 20
Solution:
(i) Cramer’s rule:
Δ = $$\left|\begin{array}{ccc} 2 & -1 & 8 \\ 3 & 4 & 5 \\ 5 & -2 & 7 \end{array}\right|$$
= 2(28 + 10) + 1(21 – 25) + 8(-6 – 20)
= 76 – 4 – 208
= -136
Δ1 = $$\left|\begin{array}{ccc} 13 & -1 & 8 \\ 18 & 4 & 5 \\ 20 & -2 & 7 \end{array}\right|$$
= 13(28 + 10) + 1(126 – 100) + 8(-36 – 80)
= 494 + 26 – 928
= -408
Δ2 = $$\left|\begin{array}{lll} 2 & 13 & 8 \\ 3 & 18 & 5 \\ 5 & 20 & 7 \end{array}\right|$$
= 2(126 – 100) – 13(21 – 25) + 8(60 – 90)
= 52 + 52 – 240
= -136
Δ3 = $$\left|\begin{array}{ccc} 2 & -1 & 13 \\ 3 & 4 & 18 \\ 5 & -2 & 20 \end{array}\right|$$
= 2(80 + 36) + 1(60 – 90) + 13(-6 – 20)
= 232 – 30 – 338
= -136 ∴ Solution is x = 3, y = 1, z = 1

(ii) Matrix inversion method:   ∴ Solution is x = 3, y = 1, z = 1

(iii) Gauss Jordan method: ∴ The given equations have a unique solution and Solution is x = 3, y = 1, z = 1. Question 7.
2x – y + 3z = 8
-x + 2y + z = 4
3x + y – 4z = 0
Solution:
(i) Cramer’s rule:
Δ = $$\left|\begin{array}{ccc} 2 & -1 & 3 \\ -1 & 2 & 1 \\ 3 & 1 & -4 \end{array}\right|$$
= 2(-8 – 1) + 1(4 – 3) + 3(-1 – 6)
= -18 + 1 – 21
= -38
Δ1 = $$\left|\begin{array}{ccc} 8 & -1 & 3 \\ 4 & 2 & 1 \\ 0 & 1 & -4 \end{array}\right|$$
= 8(-8 – 1) + 1(-16 – 0) + 3(4 – 0)
= -72 – 16 + 12
= -76
Δ2 = $$\left|\begin{array}{ccc} 2 & 8 & 3 \\ -1 & 4 & 1 \\ 3 & 0 & -4 \end{array}\right|$$
= 2(-16 – 0) – 8(4 – 3) + 3(-0 – 12)
= -32 – 8 – 36
= -76
Δ3 = $$\left|\begin{array}{ccc} 2 & -1 & 8 \\ -1 & 2 & 4 \\ 3 & 1 & 0 \end{array}\right|$$
= 2(0 – 4) + 1(0 – 12) + 8(-1 – 6)
= -8 – 12 – 56
= -76 ∴ Solution is x = 2, y = 2, z = 2.

(ii) Matrix inversion method:  Solution is x = 2, y = 2, z = 2

(iii) Gauss Jordan method:  ∴ The given equations have a unique solution and solution is x = 2, y = 2, z = 2. Question 8.
x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0
Solution:
(i) Cramer’s rule:
Δ = $$\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{array}\right|$$
= 1(-5 – 7) – 1(-2 – 14) + 1(2 – 10)
= -12 + 16 – 8
= -4
Δ1 = $$\left|\begin{array}{ccc} 9 & 1 & 1 \\ 52 & 5 & 7 \\ 0 & 1 & -1 \end{array}\right|$$
= 9(-5 – 7) – 1(-52 – 0) + 1(52 – 0)
= -108 + 52 + 52
= -4
Δ2 = $$\left|\begin{array}{ccc} 1 & 9 & 1 \\ 2 & 52 & 7 \\ 2 & 0 & -1 \end{array}\right|$$
= 1(-52 – 0) – 9(-2 – 14) + 1(0 – 104)
= -52 + 144 – 104
= -12
Δ3 = $$\left|\begin{array}{ccc} 1 & 1 & 9 \\ 2 & 5 & 52 \\ 2 & 1 & 0 \end{array}\right|$$
= 1(0 – 52) – 1(0 – 104) + 9(2 – 10)
= -52 + 104 – 72
= -20 (ii) Matrix inversion method:  Solution is x = 1, y = 3, z = 5

(iii) Gauss Jordan method: ∴ The given equations have a unique solution and solution is x = 1, y = 3, z = 5.