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