**Solve the following equations by method of reduction :**

x – 3y + z = 2 , 3x + y + z = 1 and 5x + y + 3z = 3

#### Solution

Matrix form of the given system of equations is

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

This is of the form AX = B, where

A = `[(1, -3, 1),(3, 1, 1),(5, 1, 3)], "X" = [(x),(y),(z)] "and B"= [(2),(1),(3)]`

Applying R_{2} → R_{2} – 3R_{1} and R_{3} → R_{3} – 5R_{1}, we get

`[(1, -3, 1),(0, 10, -2),(0, 16, -2)] [(x),(y),(z)] = [(2),(-5),(-7)]`

Applying R_{3} → R_{3} – `(8/5)` R_{2}, we get

`[(1, -3, 1),(0, 10, -2),(0, 0, 6/5)][(x),(y),(z)] = [(2),(-5),(1)]`

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

∴ `[(x - 3y + z),(0 + 10y - 2z),(0 + 0 + 6/5z)] = [(2),(-5),(1)]`

∴ By equality of martices, we get

x – 3y + z = 2 ...(i)

10y – 2z = – 5 ...(ii)

`(6)/(5)z` = 1

∴ z = `(5)/(6)`

Substituting z = `(5)/(6)` in equation (ii), we get

`10y - 2(5/6)` = – 5

∴ `10y - (10)/(6)` = – 5

∴ 10y = `-5 + (10)/(6) = (-20)/(6)`

∴ 10y = `(-10)/(3)`

∴ y = `(-1)/(3)`

Substituting y = `(-1)/(3)` and z = `(5)/(6)` in equation (i), we get

`x - 3((-1)/3) + (5)/(6)` = 2

∴ `x + 1 + (5)/(6)` = 2

∴ x = `2 - 1 - (5)/(6) = (1)/(6)`

∴ x = `(1)/(6)`, y = `(-1)/(3)` and z = `(5)/(6)` is the required solution.