Matrix inversion

credits.md

@@ -10,6 +10,7 @@ This work (c) 2025 is licensed under
 - Michael P. Howard
 - Evan M. Hughes
 - Michelle Schindler
+- Katie Sorensen
 
 ## Attributions
 

linear-algebra/inverses.md

@@ -1 +1,89 @@
 # Matrix inversion
+
+## 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**?
+
+```{topic} Matrix inverse
+For a square (*n* x *n*) matrix **A**, the inverse $\vv{A}^{-1}$ satisfies
+
+\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.
+
+```{topic} 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**.
+
+\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.
+
+## Inverse of a 2 x 2 matrix
+
+For a 2 x 2 matrix,
+
+\begin{equation}
+\vv{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix},
+\end{equation}
+
+the matrix inverse is
+
+\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
+
+\begin{equation}
+\vv{A} = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}
+\end{equation}
+
+First, compute its determinant:
+
+\begin{equation}
+|\vv{A}| = 3 \times 4 - 2 \times 1 = 12 - 2 = 10
+\end{equation}
+
+Then, compute its inverse
+
+\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}
+```
+
+## Inverses using Gauss-Jordan elimination
+
+For larger matrices, we can use [Gauss–Jordan elimination](gauss-jordan.md) 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 2*n* x *n* augmented matrix $[ \vv{A} \, | \, \vv{I} ]$
+
+- Perform row operations to bring to $[ \vv{I} \, | \, \vv{A}^{-1} ]$.