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Question
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
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