If #B^(-1)# is the Inverse of #B#
then
#color(white)("XXX")BxxB^(-1)=I# where #I# is the identity matrix
Suppose# B^(-1) = ((a,b),(c,d))#
Then
#color(white)("XXX")BxxB^(-1)#
#color(white)("XXX")=((-4,3),(5,2))xx((a,b),(c,d))=((1,0),(0,1))#
#rArr#
#color(white)("XXX")-4a+3c=1color(white)("XXX")#[1]
#color(white)("XXX")-4b+3d=0color(white)("XXX")#[2]
#color(white)("XXXX")5a+2c=0color(white)("XXX")#[3]
#color(white)("XXXX")5b+2d=1color(white)("XXX")#[4]
[1]#xx5 - #[3]#xx4# gives
#color(white)("XXX")23c=5 rarr c=5/23#
and substituting this back into [1] gives
#color(white)("XXX")a=-2/23#
Similarly we can use [2] and [4] to get
#color(white)("XXX")d=4/23#
#color(white)("XXX")b=2/23#
Therefore
#color(white)("XXX")B^(-1)=((-2/23,3/23),(5/23,4/23))#
