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Question
Using elementary transformations, find the inverse of the matrix A = `((8,4,3),(2,1,1),(1,2,2))`and use it to solve the following system of linear equations :
8x + 4y + 3z = 19
2x + y + z = 5
x + 2y + 2z = 7
Solution
A = IA
i.e
`[(8,4,3),(2,1,1),(1,2,2)]=[(1,0,0),(0,1,0),(0,0,1)]A`
Applying R1↔R3, we get
`[(1,2,2),(2,1,1),(8,4,3)]=[(0,0,1),(0,1,0),(1,0,0)]A`
Applying R2→R2−2R1, we get
`[(1,2,2),(0,-3,-3),(8,4,3)]=[(0,0,1),(0,1,-2),(1,0,0)]A`
Applying R3→R3−8R1, we get
`[(1,2,1),(0,-3,-3),(0,-12,-13)]=[(0,0,1),(0,1,-2),(1,0,-8)]A`
Applying R2→ `(R_2)/(−3)`, we get
`[(1,2,2),(0,1,1),(0,-12,-13)]=[(0,0,1),(0,-1/3,2/3),(1,0,-8)]A`
Applying R1→R1−2R2, we get
`[(1,0,0),(0,1,1),(0,-12,-13)]=[(0,2/3,-1/3),(0,-1/2,2/3),(1,0,-8)]A`
Applying R3→R3+12R2, we get
`[(1,0,0),(0,1,1),(0,0,-1)]=[(0,2/3,-1/3),(0,-1/3,2/3),(1,-4,0)]A`
Applying R3→−R3 and R2→R2−R3, we get
`[(1,0,0),(0,1,0),(0,0,1)]=[(0,2/3,-1/3),(1,-13/3,2/3),(-1,4,0)]A`
Thus, we have
`A^(-1)=[(0,2/3,-1/3),(1,-13/3,2/3),(-1,4,0)]`
The given system of equations is
8x+4y+3z=19
2x+y+z=5
x+2y+2z=7
This system of equations can be written as AX = B, where `A=[(8,4,3),(2,1,1),(1,2,2)],X=[(x),(y),(z)]`
∴ X=A−1B
`=>[(x),(y),(z)]=[(0,2/3,-1/3),(1,-13/3,2/3),(-1,4,0)][(19),(5),(7)]`
`=>[(x),(y),(z)][(1+10/3-7/3),(19-65/2+14/3),(-19+20+0)]=[(1),(2),(1)]`
∴x=1, y=2 and z=1
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