Question

If A = `[[1,1,1],[0,1,3],[1,-2,1]]` , find A^-1Hence, solve the system of equations: 

x +y + z = 6

y + 3z = 11

and x -2y +z = 0

Answer 1

A = `[[1,1,1],[0,1,3],[1,-2,1]]`

A₁₁ = 7 , A₁₂ = 3, A₁₃ = -1 

A₂₁ = -3 , A₂₂ = 0, A₂₃ = +3

A₃₁ = 2, A₃₂ = -3, A₃₃ = 1

|A| = 1(7) + 3 - 1= 9

`∴ A^(-1) = 1/|A|  adj A`

` = 1/9[[7,-3,2],[3,0,-3],[-1,3,1]]`

Verification 

AA^-1 = I 

`= 1/9[[1,1,1],[0,1,3],[1,-2,1]] xx  [[7,-3,2],[3,0,-3],[-1,3,1]] `

`= 1/9[[9,0,0],[0,9,0],[0,0,9]]`

=I₃

_{X +Y  + Z = 6}

_{0X + Y + 3Z = 11}

_{X -2Y + Z = 0}

`[[1,1,1],[0,1,3],[1,-2,1]] [ [X],[Y],[Z]] =[[6],[11],[0]]` 

 

aX =b ⇒ x = A^-1 b

 

`A^(-1) = 1/9 [[7,-3,2],[3,0,-3],[-1,3,1]] `

`∴ [[x],[y],[x]] =A^(-1)b`

`= 1/9 [[7,-3,2],[3,0,-3],[-1,3,1]]  [[6],[11],[0]]`

`=1/9 [[42-33],[18],[-6+33]] =1/9 [[9],[18],[27]]`

`=[[1],[2],[3]]`

∴ x =1; y =2; z = 3