Question

The sum of three numbers is 6. Thrice the third number when added to the first number, gives 7. On adding three times the first number to the sum of second and third numbers, we get 12. Find the three number by using matrices.

Answer 1

Let the numbers be x, y and z. According to the given conditions, x + y + z = 6

3z + x = 7, i.e. x + 3z = 7

and 3x + y + z = 12

Hence, the system of linear equations is

x + y + z = 6

x + 3z = 7

3x + y + z = 12

These equations can be written in matrix form as:

`[(1,1,1),(1,0,3),(3,1,1)] [("x"),("y"),("z")] = [(6),(7),(12)]`

By R₂ - R₁ and R₃ - 3R₁, we get,

`[(1,1,1),(0,-1,2),(0,-2,-2)] [("x"),("y"),("z")] = [(6),(1),(-6)]`

By R₃ + R₂, we get,

`[(1,1,1),(0,-1,2),(0,-3,0)] [("x"),("y"),("z")] = [(6),(1),(-5)]`

∴ `[("x"+"y"+"z"),(0 -"y" + "2z"),(0 -"3y" + 0)] = [(6),(1),(-5)]`

By equality of matrices,

x + y + z = 6    ...(1)

- y + 2z = 1      ...(2)

- 3y = - 5        ...(3)

From (3), y = `5/3`

Substituting y = `5/3` in (2), we get,

`- 5/3 + 2"z" = 1`

∴ 2z = `1 + 5/3 = 8/3`

∴ z = `4/3`

Substituting y = `5/3`, y = `4/3` in (1), we get,

`"x" + 5/3 + 4/3 = 6`

∴ x = 3

∴ x = 3, y = `5/3`, z = `4/3`

Hence, the required numbers are `3, 5/3` and `4/3`.