Inverse of a Matrix

Definition §

A $n\times n$ matrix is invertible if there is an $n\times n$ matrix $\mathbf{C}$ such that $\mathbf{CA}=\mathbf{I}$ and $\mathbf{AC}=\mathbf{I}$, where $\mathbf{I}$ is a $n\times n$ identity matrix.

Finding the Inverse §

Row Operations §

We can use the product of elementary matrices to find the inverse of a matrix. Elementary matrices are matrices that can be derived from the identity matrix with one step.

Example §

Find the inverse of $\mathbf{A}$

$$\begin{array}{rl}A& =\left(\begin{array}{ccc}4& 0& 5\\ 0& 3& 0\\ 1& 0& 4\end{array}\right)\\ E_1=\left(\begin{array}{ccc}\frac{1}{4}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\to A& =\left(\begin{array}{ccc}1& 0& \frac{5}{4}\\ 0& 3& 0\\ 1& 0& 4\end{array}\right)\\ E_2=\left(\begin{array}{ccc}1& 0& 0\\ 0& \frac{1}{3}& 0\\ 0& 0& 1\end{array}\right)\to A& =\left(\begin{array}{ccc}1& 0& \frac{5}{4}\\ 0& 1& 0\\ 1& 0& 4\end{array}\right)\\ E_3=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ -1& 0& 1\end{array}\right)\to A& =\left(\begin{array}{ccc}1& 0& \frac{5}{4}\\ 0& 1& 0\\ 0& 0& 4\end{array}\right)\\ E_4=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ -1& 0& 1\end{array}\right)\to A& =\left(\begin{array}{ccc}1& 0& \frac{5}{4}\\ 0& 1& 0\\ 0& 0& \frac{11}{4}\end{array}\right)\\ E_5=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& \frac{4}{11}\end{array}\right)\to A& =\left(\begin{array}{ccc}1& 0& \frac{5}{4}\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\\ E_6=\left(\begin{array}{ccc}1& 0& -\frac{5}{4}\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\to A& =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\\ A^{-1}=E_1{E}_2{E}_3{E}_4{E}_5{E}_6& \end{array}$$

Adjugate Matrix §

The inverse of a matrix $\mathbf{A}$ in terms of the determinant and its adjugate matrix is

$${\mathbf{A}}^{-1}=\frac{1}{\det \mathbf{A}}adj(\mathbf{A})$$

Example §

$$\begin{array}{rl}adj(\mathbf{A})& =\left(\begin{array}{ccc}2& 1& 3\\ 1& -1& 1\\ 1& 4& -2\end{array}\right)\\ & ={\left(\begin{array}{ccc}-2& 3& 5\\ 14& -7& -7\\ 4& 1& -3\end{array}\right)}^T\\ & =\left(\begin{array}{ccc}-2& 14& 4\\ 3& -7& 1\\ 5& -7& -3\end{array}\right)\\ \det \mathbf{A}& =2\cdot -2+1\cdot 3+3\cdot 5\\ & =14\\ {\mathbf{A}}^{-1}& =\frac{1}{14}\left(\begin{array}{ccc}-2& 14& 4\\ 3& -7& 1\\ 5& -7& -3\end{array}\right)\\ & =\left(\begin{array}{ccc}-\frac{1}{7}& 1& \frac{2}{7}\\ \frac{3}{14}& -\frac{1}{2}& \frac{1}{14}\\ \frac{5}{14}& -\frac{1}{2}& -\frac{3}{14}\end{array}\right)\end{array}$$

Properties §

Theorem 1 §

$$(\mathbf{AB}{)}^{-1}={\mathbf{B}}^{-1}\cdot {\mathbf{A}}^{-1}$$

Theorem 2 §

Given Theorem 1, we can use an inductive proof to show that given that $A_1,A_2,A_3\dots A_n$ are invertible matrices,

$$(A_1\cdot A_2\cdot A_3\dots A_n{)}^{-1}=A_n^{-1}\cdot A_{n-1}^{-1}\dots A_1^{-1}$$

Base case: n = 2

The theorem literally states this is true

Inductive step:

Let $A=(A_1\cdot A_2\dots A_n)$, $B=A_{n+1}$

$$\begin{array}{rl}(A_1\cdot A_2\dots A_{n+1}{)}^{-1}& =(A\cdot B{)}^{-1}\\ & =B^{-1}\cdot A^{-1}\\ & =A_{n+1}^{-1}\cdot A_n^{-1}\cdot A_{n-1}^{-1}\dots A_1\end{array}$$

$S(n+1)$ holds true given $S(n)$ is true.