How do you find the inverse of #A=##((6, 7, 8), (1, 0, 1), (0, 1, 0))#?

1 Answer
Dec 17, 2016

Answer:

The answer is # =((-1/2,4,7/2),(0,0,1),(1/2,-3,-7/2))#

Explanation:

First we calculate the determinant of matrix A

#detA= | (6,7,8), (1,0,1), (0,1,0) | #

#=6(0-1)-7(0)+8(1)=2#

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

Then, we calculate the matrix of minor cofactors

#A^c= ( (((0,1),(1,0)),-((1,1),(0,0)),((1,0),(0,1))), (-((7,8),(1,0)),((6,8),(0,0)),-((6,7),(0,1))), (((7,8),(0,1)),-((6,8),(1,1)),((6,7),(1,0))) ) #

#= ( (-1,0,1), (8,0,-6), (7,2,-7) )#

Then we calculate the transpose of #A^c#

#barA^c=((-1,8,7),(0,0,2),(1,-6,-7))#

Then, the inverse is

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

#=1/2((-1,8,7),(0,0,2),(1,-6,-7))#

#=((-1/2,4,7/2),(0,0,1),(1/2,-3,-7/2))#

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

#=((6,7,8), (1,0,1), (0,1,0) )*((-1/2,4,7/2),(0,0,1),(1/2,-3,-7/2)) #

#=((1,00),(0,10),(0,0,1))#

#=I#