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Matrix Inversion

Module by: Thanos Antoulas, JP Slavinsky, Bailey Edwards

Summary: (Blank Abstract)

Let's say we have the square nn x nn matrix AA composed of real numbers. By "square", we mean it has the same number of rows as columns.

A=a11…a1n⋮⋱⋮an1…ann A a11 … a1n ⋮ ⋱ ⋮ an1 … ann (1)

The subscripts of the real numbers in this matrix denote the row and column numbers, respectively (i.e. a12a12 holds the position at the intersection of the first row and the second column).

We will denote the inverse of this matrix as A-1 A . A matrix inverse has the property that when it is multiplied by the original matrix (on the left or on the right), the result will be the identity matrix.

AA-1=A-1A=I A A A A I (2)

To compute the inverse of AA, two steps are required. Both involve taking determinants ( detA A ) of matrices. The first step is to find the adjoint ( AH A ) of the matrix A. It is computed as follows:

AH=α11…α1n⋮⋱⋮αn1…αnn A α11 … α1n ⋮ ⋱ ⋮ αn1 … αnn (3)
αij=-1i+jdet A ij αij -1 i j A ij (4)

where A ij A ij is the n-1n1 x n-1n1 matrix obtained from AA by eliminating its ii-th column and jj-th row. Note that we are not eliminating the ii-th row and jj-th column as you might expect.

To finish the process of determining the inverse, simply divide the adjoint by the determinant of the original matrix AA .

A-1=1detAAH A 1 A A (5)

Example 1

A=abcd A a b c d (6)

Find the inverse of the above matrix.

The first step is to compute the terms in the adjoint matrix:

α11=-12d α11 -1 2 d (7)
α12=-13b α12 -1 3 b (8)
α21=-13c α21 -1 3 c (9)
α22=-14a α22 -1 4 a (10)

Therefore,

AH=d-b-ca A d b c a (11)

We then compute the determinant to be ad-bc a d b c . Dividing through by this quantity yields the inverse of AA:

A-1=1ad-bcd-b-ca A 1 a d b c d b c a (12)

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