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

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 matrix

For #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 method

Use 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)]#.