Question

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

2x – 2y + 3z = 2, x + 2y – z = 3, 3x – y + 2z = 1

Answer 1

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