# 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 - Mathematics

Sum

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

#### Solution

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

Is there an error in this question or solution?