Inverse matrix of 2-by-2 matrix, 3-by-3 matrix, 4-by-4 matrix

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Inverse matrices of 2-by-2 matrix, 3-by-3 matrix, 4-by-4 matrix are shown here.


Inverse matrix of 2 $\times$ 2 matrix

A=abcd

There exists an inverse matrix of A when detA=ad-bc0 , and it is

A-1=1ad-bcd-b-ca

Inverse matrix of 3 $\times$ 3 matrix

A=a11a12a13a21a22a23a31a32a33

There exists an inverse matrix of A when
detA=a11a22a33+a12a23a31+a13a21a32-a13a22a31-a11a23a32-a12a21a33
0 , and it is

A-1=1detAa22a33-a23a32a13a32-a12a33a12a23-a13a22a23a31-a21a33a11a33-a13a31a13a21-a11a23a21a32-a22a31a12a31-a11a32a11a22-a12a21

Inverse matrix of 4 $\times$ 4 matrix

A=a11a12a13a14a21a22a23a24a31a32a33a34a41a42a43a44

If

detA
=a11a22a33a44+a11a23a34a42+a11a24a32a43
+a12a21a34a43+a12a23a31a44+a12a24a33a41
+a13a21a32a44+a13a22a34a41+a13a24a31a42
+a14a21a33a42+a14a22a31a43+a14a23a32a41
-a11a22a34a43-a11a23a32a44-a11a24a33a42
-a12a21a33a44-a12a23a34a41-a12a24a31a43
-a13a21a34a42-a13a22a31a44-a13a24a32a41
-a14a21a32a43-a14a22a33a41-a14a23a31a42
0

then there exists an inverse matrix of A , and it is

A-1=1detAb11b12b13b14b21b22b23b24b31b32b33b34b41b42b43b44

where

b11=a22a33a44+a23a34a42+a24a32a43-a22a34a43-a23a32a44-a24a33a42
b12=a12a34a43+a13a32a44+a14a33a42-a12a33a44-a13a34a42-a14a32a43
b13=a12a23a44+a13a24a42+a14a22a43-a12a24a43-a13a22a44-a14a23a42
b14=a12a24a33+a13a22a34+a14a23a32-a12a23a34-a13a24a32-a14a22a33
b21=a21a34a43+a23a31a44+a24a33a41-a21a33a44-a23a34a41-a24a31a43
b22=a11a33a44+a13a34a41+a14a31a43-a11a34a43-a13a31a44-a14a33a41
b23=a11a24a43+a13a21a44+a14a23a41-a11a23a44-a13a24a41-a14a21a43
b24=a11a23a34+a13a24a31+a14a21a33-a11a24a33-a13a21a34-a14a23a31
b31=a21a32a44+a22a34a41+a24a31a42-a21a34a42-a22a31a44-a24a32a41
b32=a11a34a42+a12a31a44+a14a32a41-a11a32a44-a12a34a41-a14a31a42
b33=a11a22a44+a12a24a41+a14a21a42-a11a24a42-a12a21a44-a14a22a41
b34=a11a24a32+a12a21a34+a14a22a31-a11a22a34-a12a24a31-a14a21a32
b41=a21a33a42+a22a31a43+a23a32a41-a21a32a43-a22a33a41-a23a31a42
b42=a11a32a43+a12a33a41+a13a31a42-a11a33a42-a12a31a43-a13a32a41
b43=a11a23a42+a12a21a43+a13a22a41-a11a22a43-a12a23a41-a13a21a42
b44=a11a22a33+a12a23a31+a13a21a32-a11a23a32-a12a21a33-a13a22a31

Inverse matrix of NxN matrix

From the analogy of the above formulae, the computation time of inverse matrix of NxN matrix will be O(N3N!). Computing inverse matrix with Gauss-Jordan method, the method using LU decomposition, and the method using SVD, will take a computation time of O(N3) (not confident). I will recommend not to use the formula for calculating inverse matrix of NxN matrix which N >= 4.


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