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Question
Solve the following systems of linear equations by Gaussian elimination method:
2x – 2y + 3z = 2, x + 2y – z = 3, 3x – y + 2z = 1
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Solution
The matrix form of the above equations is
`[(2, -2, 3),(1, 2, -1),(3, -1, 2)][(x),(y),(z)] = [(2),(3),(1)]`
(i.e) AX = B
The augment matrix (A, B) is
(A, B) = `[(2, -2, 3, 2),(1, 2, -1, 3),(3, -1, 2, 1)] ˜ [(1, 2, -1, 3),(2, -2, 3, 2),(3, -1, 2, 1)] "R"_1 ↔ "R"_2`
` ˜ [(1, 2, -1, 3),(0, -6, 5, -4),(0, -7, 5, -8)] "R"_2 -> "R"_2 - 2"R"_1; "R"_3 -> "R"_3 - 3"R"_1`
` ˜ [(1, 2, -1, 3),(0, -6, 5, -4),(0, 0, -5, -20)] "R"_3 -> 6"R"_3 - 7R"_2`
The above matrix is in echelon form.
Now writing the equivalent equations
`[(1, 2, -1),(0, -6, 5),(0, 0, -5)][(x),(y),(z)] = [(3),(-4),(-20)]`
(i.e) x + 2 – z = 3
– 6y + 5z = – 4
– 5z = – 20
From (3)
⇒ z = `(-20)/(-5)` = 4
Substituting z = 4 in (2) we get
– 6y + 20 = – 4
⇒ – 6y = – 4 – 20 = – 24
⇒ y = 4
Substituting z = 4 and y = 4 in (1) we get
x + 8 – 4 = 3
⇒ x + 4 = 3
⇒ x = 3 – 4 = – 1
So, x = – 1, y = 4, z = 4
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