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

1 Answer
Dec 17, 2016

Answer:

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#