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

##### 1 Answer

#### Answer:

#### 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 matrix**

For

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

**2) Augmented matrix method**

Use Gauss-Jordan elimination to transform

The following steps result in

#[(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) Adjoint method**

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

The adjoint of

The

So for a

#"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