## Question

###### How do you find the inverse of #A=##((3, 5), (2, 4))#?

## Answers

#[(3,5),(2,4)]^{-1} = [(2,-5/2),(-1,3/2)]# ## Explanation:

Find its determinant first.

#det(A) = 3xx4-5xx2 = 2# From here, there are many "different" ways of finding inverse matrix, here are 3 of them:

1) Shortcut for 2x2 matrixFor

#A = [(a,b),(c,d)]# , the inverse can be found using the formula:

#A^{-1} = frac{1}{color(green)(det(A))}[(d,-b),(-c,a)] = frac{1}{color(green)(ad-bc)}[(d,-b),(-c,a)]# So for

#A = [(3,5),(2,4)]# ,

#A^{-1} = 1/2 [(4,-5),(-2,3)] = [(2,-5/2),(-1,3/2)]#

2) Augmented matrix methodUse Gauss-Jordan elimination to transform

#[A|I]# to#[I|A^{-1}]# .The following steps result in

#[(3,5),(2,4)]^{-1}# .

#[(3,5|1,0),(2,4|0,1)] -> [(6,10|2,0),(6,12|0,3)]#

#-> [(6,10|2,0),(0,2|-2,3)]#

#-> [(6,0|12,-15),(0,2|-2,3)]#

#-> [(1,0|2,-5/2),(0,1|-1,3/2)]# So we see that

#[(3,5),(2,4)]^{-1} = [(2,-5/2),(-1,3/2)]# .

3) Adjoint method

#A^{-1} = 1/(det(A))"adj"(A)# The adjoint of

#nxxn# matrix#A# is another#nxxn# matrix in which the#(i,j)# entry is the#(j,i)# cofactor of#A# .The

#(i,j)# cofactor of A is the determinant of the#(n - 1)xx(n - 1)# matrix that results from deleting row#i# and column#j# of#A# , multiplied by#(-1)^((i+j))# . This means that you multiply the determinant by#-1# if#(i+j)# is odd and you don't do anything if#(i+j)# is even.So for a

#2xx2# matrix, let#A = [(a,b),(c,d)]# , then the adjoint is given by

#"adj"(A) = [(det([d]),-det([c])),(-det([b]),det([a]))]^T#

#= [(d,-c),(-b,a)]^T#

#= [(d,-b),(-c,a)]# So,

#A^{-1} = 1/(det(A))[(d,-b),(-c,a)]# This is the "shortcut" we used in method

1).To check your answer, just compute

#[(3,5),(2,4)][(2,-5/2),(-1,3/2)]# and#[(2,-5/2),(-1,3/2)][(3,5),(2,4)]# Both should give

#[(1,0),(0,1)]# .