Express the following equations in the matrix form and solve them by method of reduction :

2x- y + z = 1, x + 2y + 3z = 8, 3x + y - 4z =1

#### Solution

The matrix form of given equations is

`[[2,-1,1],[1,2,3],[3,1,-4]][[x],[y],[z]]=[[1],[8],[1]]`

`R_1 harr R_2`

`[[1,2,3],[2,-1,1],[3,1,-4]][[x],[y],[z]]=[[8],[1],[1]]`

`R_2-> R_2+R_1`

`[[1,2,3],[3,1,4],[3,1,-4]][[x],[y],[z]]=[[8],[9],[1]]`

`R_3 -> R_3-R_2`

`[[1,2,3],[3,1,4],[0,0,-8]][[x],[y],[z]]=[[8],[9],[-8]]`

`R_2->R_2-3R_1`

`[[1,2,3],[0,-5,-5],[0,0,-8]][[x],[y],[z]]=[[8],[-15],[-8]]`

`[[x+2y+3z],[-5y-5z],[-8z]]=[[8],[-15],[-8]]`

therefore

x + 2y + 3z = 8 .............(1)

-5y -5z = -15....... (2)

-8z = -8..............(3)

From (3),

**z = 1**

From (2),

-5y - 5(1) = -15 ... (because z = 1)

-5y =-10

**y = 2**

From (1),

x + 2(2)+ 3(1) = 8 ... (because z = 1 and y = 2)

x = 8 -7

**x = 1**

**Thus, x = 1, y = 2, z = 1**