The sum of three numbers is 6. If we multiply third number by 3 and add it to the second number we get 11. By adding the first and third number we get a number which is double the second number. Use this information and find a system of linear equations. Find the three numbers using matrices.

#### Solution

Let the three numbers be x, y and z respectively.

According to the first condition,

x + y + z = 6

According to the second condition,

3z + y = 11 i.e., y + 3z = 11

According to the third condition,

x + z = 2y i.e., x – 2y + z = 0

Matrix form of the above system of equations is

`[(1, 1, 1),(0, 1, 3),(1, -2, 1)] [(x),(y),(z)] = [(6),(11),(0)]`

Applying R_{2} ↔ R_{3}, we get

`[(1, 1, 1),(1, -2, 1),(0, 1, 3)] [(x),(y),(z)] = [(6),(0),(11)]`

Applying R_{2} → R_{2} – R_{1}, we get

`[(1, 1, 1),(0, -3, 0),(0, 1, 3)] [(x),(y),(z)] = [(6),(-6),(11)]`

Applying R_{3} → 3R_{3} + R_{2}, we get

`[(1, 1, 1),(0, -3, 0),(0, 0, 9)] [(x),(y),(z)] = [(6),(-6),(27)]`

Hence, the original matrix is reduced to an upper triangular matrix.

∴ `[(x + y + z),(0 - 3y + 0),(0 + 0 + 9z)][(6),(-6),(27)]`

∴ By equality of martices, we get

x + y + z = 6 ......(i)

– 3y = – 6 ......(ii)

i.e., y = 2

9z = 27 ......(iii)

i.e., z = 3

Substituting y = 2 and z = 3 in equation (i), we get

x + 2 + 3 = 6

∴ x = 1

∴ 1, 2 and 3 are the required numbers.