Question

How do you find the inverse of `A=``((-5, 1, 4), (-1, 4, 5), (-1, 0, -1))`?

Answer 1

The answer is `=((-2/15,1/30,-11/30),(-1/5,3/10,7/10),(2/15,-1/30,-19/30))`

Explanation:

First we calculate the determinant of matrix A

`detA= | (-5,1,4), (-1,4,5), (-1,0,-1) |`

`=-5(0-4)-1(1+5)+4(4)=20-6+16=30`

As `detA!=0`, the matrix is invertible

Then, we calculate the matrix of minor cofactors

`A^c= ( (((4,5),(0,-1)),-((-1,5),(-1,-1)),((-1,4),(-1,0))), (-((1,4),(0,-1)),((-5,4),(-1,-1)),-((-5,1),(-1,0))), (((1,4),(4,5)),-((-5,4),(-1,5)),((-5,1),(-1,4))) )`

`= ( (-4,-6,4), (1,9,-1), (-11,21,-19) )`

Then we calculate the transpose of `A^c`

`barA^c=((-4,1,-11),(-6,9,21),(4,-1,-19))`

Then, the inverse is

`A^(-1)=barA^c/detA`

`=1/30((-4,1,-11),(-6,9,21),(4,-1,-19))`

`=((-2/15,1/30,-11/30),(-1/5,3/10,7/10),(2/15,-1/30,-19/30))`

Verification, by doing `A A^(-1)`

`=((-5,1,4), (-1,4,5), (-1,0,-1) )*((-2/15,1/30,-11/30),(-1/5,3/10,7/10),(2/15,-1/30,-19/30))`

`=((1,0,0),(0,1,0),(0,0,1))`

`=I`