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

Δ_{1} = \(\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

Δ_{2} = \(\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

Δ_{3} = \(\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

Δ_{1} = \(\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

Δ_{2} = \(\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

Δ_{3} = \(\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]\)

R_{2} → R_{2} – 2R_{1}, R_{3} → R_{3} – R_{1}

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

Δ_{1} = \(\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

Δ_{2} = \(\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

Δ_{3} = \(\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

Δ_{1} = \(\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

Δ_{2} = \(\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

Δ_{3} = \(\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

Δ_{1} = \(\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

Δ_{2} = \(\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

Δ_{3} = \(\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

Δ_{1} = \(\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

Δ_{2} = \(\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

Δ_{3} = \(\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

Δ_{1} = \(\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

Δ_{2} = \(\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

Δ_{3} = \(\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.