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Question
Solve the following equations by reduction method :
x + 2y + z = 8
2x+ 3y - z = 11
3x - y - 2z = 5
Solution
Given :
x + 2y + z = 8
2x+ 3y - z = 11
3x - y - 2z = 5
Its matrix form is
`[(1,2,1),(2,3,-1),(3,-1,-2)] [(x),(y),(z)] = [(8),(11),(5)]`
R2 → R2 - 2R1 , R3 → R3 - 3R1
`[(1,2,1),(0,-1,-3),(0,-7,-5)] [(x),(y),(z)] = [(8),(-5),(-19)]`
R3 → R3 - 7 R2
`[(1,2,1),(0,-1,-3),(0,0,16)] [(x),(y),(z)] = [(8),(-5),(16)]`
Now write the equation in the original form
x + 2y + z = 8 ... (i)
- y- 3z = -5 ... (ii)
I6z = 16 .....(iii)
From (iii). z = 1
Putting z = 1 in equation (ii)
We get y = 2
Putting z = 1, y = 2 in equation (i)
we get x = 3
∴ Solution is x = 3, y = 2, z = 1.
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