Inter 1st Year Maths 1A Matrices Solutions Ex 3(h)

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.