Question

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

Answer 1

`[(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 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)]`.