How do you find the inverse of #A=##((1, 2, 0), (-2, -2, 1), (0, -1, -1))#?

1 Answer
Oct 25, 2016

Answer:

The inverse is

#A^(-1)=((-3,-2,-2),(2,1,1),(-2,-1,-2))#

Explanation:

We start by calculating the determinant of A
#detA=1*(2+1)-2*(2)*0=3-4=-1#
The determinant of A is #!=0# , so the matrix has an inverse.

#A=((1,2,0),(-2,-2,1),(0,-1,-1))#

Now we calculate the 9 minors and cofactors

#M_11=3##color(white)(aaaaaaa)# #C_11=3#
#M_21=-2##color(white)(aaaaa)# #C_21=2#
#M_31=2##color(white)(aaaaaaa)# #C_31=2#
#M_12=2##color(white)(aaaaaaa)# #C_12=-2#
#M_22=-1##color(white)(aaaaa)# #C_22=-1#
#M_32=1##color(white)(aaaaaaa)# #C_32=-1#
#M_13=2##color(white)(aaaaaaa)# #C_13=2#
#M_23=-1##color(white)(aaaaaa)# #C_23=-1#
#M_33=2##color(white)(aaaaaaa)# #C_33=2#

matrix of co factors = #((3,-2,2),(2,-1,1),(2,-1,2))#

Then the transpose = #((3,2,2),(-2,-1,-1),(2,1,2))#

Then we divide by the determinant to obtain the inverse

=#((-3,-2,-2),(2,1,1),(-2,-1,-2))#

Check by mltiplying the matrix and its inverse

#((-3,-2,-2),(2,1,1),(-2,-1,-2))##((1,2,0),(-2,-2,1),(0,-1,-1))#*#=((1,0,0),(0,1,0),(0,0,1))#