3.5. Matrix inversion

3.5.1. Motivation and definition

Gauss-Jordan elimination works well for solving Ax = b, but the process needs to be repeated for every new b. Is there an alternative if we need to solve Ax = b for many different b?

Matrix inverse

For a square (n x n) matrix A, the inverse \(\vv{A}^{-1}\) satisfies

(3.24)\[\begin{equation} \vv{A} \vv{A}^{-1} = \vv{A}^{-1} \vv{A} = \vv{I} \end{equation}\]

where I is the n x n identity matrix.

A matrix is called nonsingular or invertible if it has an inverse, but singular if it does not.

Invertible matrix theorem

A is invertible if and only if the determinant of A is nonzero.

(There are many more such conditions!)

If the inverse of A exists, it is unique and can be used to solve Ax = b.

(3.25)\[\begin{align} \vv{A} \vv{x} &= \vv{b} \\ \vv{A}^{-1} \vv{A} \vv{x} &= \vv{A}^{-1} \vv{b} \\ \vv{x} &= \vv{A}^{-1} \vv{b} \end{align}\]

Finding the inverse of A is usually hard. There is a general definition based on cofactors, as well as advanced numerical methods, that we will not cover. Instead, we focus on two options: a formula for 2 x 2 matrices, and use of Gauss-Jordan elimination for larger matrices.

3.5.2. Inverse of a 2 x 2 matrix

For a 2 x 2 matrix,

(3.26)\[\begin{equation} \vv{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \end{equation}\]

the matrix inverse is

(3.27)\[\begin{equation} \vv{A}^{-1} = \frac{1}{|\vv{A}|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \end{equation}\]

(Flip a and d, change the signs of b and c.)

Example: 2 x 2 inverse

To find the inverse of

(3.28)\[\begin{equation} \vv{A} = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} \end{equation}\]

First, compute its determinant:

(3.29)\[\begin{equation} |\vv{A}| = 3 \times 4 - 2 \times 1 = 12 - 2 = 10 \end{equation}\]

Then, compute its inverse

(3.30)\[\begin{equation} \vv{A}^{-1} = \frac{1}{10} \begin{bmatrix} 4 & -1 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 0.4 & -0.1 \\ -0.2 & 0.3 \end{bmatrix} \end{equation}\]

3.5.3. Inverses using Gauss-Jordan elimination

For larger matrices, we can use Gauss–Jordan elimination to solve \(\vv{A} \vv{A}^{-1} = \vv{I}\) as a generalization of Ax = b.

• Check that \(|\vv{A}| \ne 0\) (i.e., A is invertible).

• Form the 2n x n augmented matrix \([ \vv{A} \, | \, \vv{I} ]\)

• Perform row operations to bring to \([ \vv{I} \, | \, \vv{A}^{-1} ]\).