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

1 Answer
Dec 8, 2016

Answer:

#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))#