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प्रश्न
Solve the following equations by reduclion method
x+3y+3z= 16 , x+4y+4z=21 , x+3y+4z = 19
उत्तर
Matrix equation is
`[(1,3,3),(1,4,4),(1,3,4)][(x),(y),(z)] = [(16),(21),(19)]`
R2 → R2 - R3
`[(1,3,3),(0,1,0),(1,3,4)] [(x),(y),(z)] = [(16),(2),(19)]`
R3 → R3 - R1
`[(1,3,3),(0,1,0),(0,0,1)][(x),(y),(z)] = [(16),(2),(3)]`
`[(x+3y+3z),(0+y+0),(0+0+z)] = [(16),(2),(3)]`
∴ x+3y+3z = 16
y = 2
z = 3
∴ x + 3(2) + 3(3) = 16
∴ x + 6 + 9 = 16
∴ x = 16 - 15 = 1
∴ x = 1 , y = 2 , z = 3
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