How do you find the inverse of #[(3/4, -1/8), (5,1/2)]#?

1 Answer
Dec 2, 2016

Answer:

The inverse is #=( (1/2,1/8), (-5,3/4) )#

Explanation:

let the matrix be #A= ( (3/4,-1/8), (5,1/2) )#

We calculate the determinant

#det(A)=| (3/4,-1/8), (5,1/2) | #

#=3/4*1/2-(-1/8*5)=3/8+5/8=1!=0#

For a matrix to be invertible, the determinant #!=0#

Now we calculate the matrix of the minor cofactor

#A_c=##( (1/2,-5), (1/8,3/4) )#

Now we calculate the transpose of #A_c^t#

#=( (1/2,1/8), (-5,3/4) )#

and the inverse is

#A^(-1)=1/detA( (1/2,1/8), (-5,3/4) )#

#=( (1/2,1/8), (-5,3/4) )#

Verification

#A*A^(-1)= ( (3/4,-1/8), (5,1/2) )*( (1/2,1/8), (-5,3/4) )#

#( (1,0), (0,1) )=I#