Solve the following systems of linear equations by Gaussian elimination method: 2x + 4y + 6z = 22, 3x + 8y + 5z = 27, – x + y + 2z = 2

Question

Solve the following systems of linear equations by Gaussian elimination method:

2x + 4y + 6z = 22, 3x + 8y + 5z = 27, – x + y + 2z = 2

Sum

Solution

2x + 4y + 6z = 22 .......(1)

3x + 8y + 5z = 27 .......(2)

– x + y + 2z = 2 .......(3)

Divide equation (1) by 2 we get

x + 2y + 3z = 11 .......(1)

3x + 8y + 5z = 27 .......(2)

– x + y + 2z = 2 .......(3)

The matrix form of the above equations is

`[(1, 2, 3),(3, 8, 5),(-1, 1, 2)][(x),(y),(z)] = [(11),(27),(2)]`

(i.e) AX = B

The augment matrix (A, B) is

(A, B) = `[(1, 2, 3, 11),(3, 8, 5, 27),(-1, 1, 2, 2)]`

`˜ [(1, 2, 3, 11),(0, 2, -4, -6),(0, 3, 5, 13)] "R"_2 -> "R"_2 - 3"R"_1 ; "R"_3 -> "R"_3 + "R"_1`

`˜ [(1, 2, 3, 11),(0, 1, -2, -3),(0, 3, 5, 13)] "R"_2 -> "R"_2/2`

`˜ [(1, 2, 3, 11),(0, 1, -2, -3),(0, 0, 11, 22)] "R"_3 -> "R"_3 - 3"R"_2`

The above matrix is in echelon form.

Now writing the equivalent equations.

`[(1, 2, 3),(0, 1, -2),(0, 0, 11)][(x),(y),(z)] = [(11),(-3),(22)]`

⇒ x + 2y + 3z = 11

y – 2z = – 3

11z = 22

From (3)

⇒ z = `22/11`

= 2

Substituting z = 2 in (2) we get

y – 4 = – 3

⇒ y = – 3 + 4 = 1

Substituting z = 2, y = 1 in (1) we get

x + 2(1) + 3(2) = 11

⇒ x + 2 + 6 = 11

⇒ x + 8 = 11

⇒ x = 11 – 8 = 3

x = 3, y = 1, z = 2

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Applications of Matrices: Solving System of Linear Equations

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Q 1. (ii)Q 1. (i)Q 2

Chapter 1: Applications of Matrices and Determinants - Exercise 1.5 [Page 37]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board

Chapter 1 Applications of Matrices and Determinants
Exercise 1.5 | Q 1. (ii) | Page 37

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