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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 - Mathematics

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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

2xyz = 5

x + 2y + 2z = 7

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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 R1R3, we get

`[(1,2,2),(2,1,1),(8,4,3)]=[(0,0,1),(0,1,0),(1,0,0)]A`

Applying R2R22R1, we get

`[(1,2,2),(0,-3,-3),(8,4,3)]=[(0,0,1),(0,1,-2),(1,0,0)]A`

Applying R3R38R1, 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 R1R12R2, 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 R3R3+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 R3R3 and R2R2R3, 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=A1B

`=>[(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

Concept: Elementary Transformations
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