Question

How do you find the inverse of `A=``((2, 3, 1), (4, 3, 1), (1, 2, 4))`?

Answer 1

`A^(-1) = 1/20((-10, 10, 0),(15, -7, -2),(-5,1,6))`

Explanation:

In place of each element of `A=((2, 3, 1),(4, 3, 1),(1, 2, 4))` write down the determinant of the `2xx2` matrix formed by deleting the row and column that the current element is in:

`((abs((3,1),(2,4)),abs((4,1),(1,4)),abs((4,3),(1,2))), (abs((3,1),(2,4)), abs((2,1),(1,4)), abs((2,3),(1,2))), (abs((3,1),(3,1)), abs((2,1),(4,1)), abs((2,3),(4,3)))) = ((10,15,5),(10,7,1),(0,-2,-6))`

Invert the signs of alternate elements in a checkboard pattern:

`((+, -, +), (-, +, -), (+, -, +))`

to get:

`((10,-15,5),(-10,7,-1),(0,2,-6))`

Transpose to get:

`((10, -10, 0),(-15, 7, 2),(5,-1,-6))`

This will be a scalar multiple of the inverse matrix - the multiplier being `abs(A)`.

So we find:

`((2, 3, 1),(4, 3, 1),(1, 2, 4))((10, -10, 0),(-15, 7, 2),(5,-1,-6)) = ((-20,0,0),(0,-20,0),(0,0,-20))`

Hence:

`A^(-1) = -1/20((10, -10, 0),(-15, 7, 2),(5,-1,-6)) = 1/20((-10, 10, 0),(15, -7, -2),(-5,1,6))`