ML Wiki

Inverse Matrices

A square $n \times n$ matrix $A$ has inverse (or $A$ is invertible) if there exists $B$ s.t. $A \times B = B \times A = I_n$

• If $B$ exists, then it's denoted $A^{-1}$
• $A$ in such case is called non-singular
• otherwise (no $A^{-1}$ exists) $A$ is called singular

There are two types of inverses:

• left and right
• $\underbrace{A \times A^{-1}}_\text{left} = I_n = \underbrace{A^{-1} \times A}_\text{right}$
• for square matrices left and right inverses are equal

Finding the Inverse

Gauss-Jordan Elimination

Suppose we have an equation $A \times A^{-1} = I$

• how can we solve it to find $A^{-1}$? Let's replace $A^{-1}$ by $X$ and solve $A \times X = I$
• $A \times X = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} \times \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$
• one idea: Solve $n$ different systems of linear equations
• $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} \times \begin{bmatrix} x_{11} \\ x_{21} \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and
• $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} \times \begin{bmatrix} x_{12} \\ x_{22} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$
• i.e. for $i$th system, take $i$th column of $X$ ($\mathbf x_i$) and $i$th row of $I$ ($\mathbf e_i$)
• we have a bunch of systems like $A \mathbf x_i = \mathbf e_i$ that we know how to solve
• so we can use Gaussian Elimination for that
• we'll have several augmented matrices like $\left[ \begin{array}{cc|c} a_{11} & a_{12} & 1 \\ a_{21} & a_{22} & 0 \\ \end{array} \right]$ and $\left[ \begin{array}{cc|c} a_{11} & a_{12} & 0 \\ a_{21} & a_{22} & 1 \\ \end{array} \right]$ that we can solve to get $\begin{bmatrix} x_{11} \\ x_{21} \\ \end{bmatrix}$ and $\begin{bmatrix} x_{12} \\ x_{22} \\ \end{bmatrix}$
• but we can also put all such vectors $\mathbf x_i$ and $\mathbf e_i$ at the same time!
• $\left[ \begin{array}{cc|cc} a_{11} & a_{12} & 1 & 0 \\ a_{21} & a_{22} & 0 & 1 \\ \end{array} \right]$

Gaussian Elimination:

• so once we have an augmented matrix $\Big[ \ A \; \Big| \; I \ \Big] = \left[ \begin{array}{cc|cc} a_{11} & a_{12} & 1 & 0 \\ a_{21} & a_{22} & 0 & 1 \\ \end{array} \right]$
• we come from $A$ to $I$ while applying the same actions to the augmented part $I$.
• at the end we should get $\Big[ \ A \; \Big| \; I \ \Big] \to \Big[ \ I \; \Big| \; A^{-1} \ \Big]$

Why does it work?

• suppose you did your elimination on $A$ alone, so you obtained $EA = I$ (assume no row exchanges)
• let's apply $E$ to augmented $\Big[ \ A \; \Big| \; I \ \Big]$.
• $E \times \Big[ \ A \; \Big| \; I \ \Big] = \Big[ \ EA \; \Big| \; EI \ \Big] = \Big[ \ I \; \Big| \; E \ \Big]$
• what is $E$? Since $EA = I$ we know that it can be only when $E = A^{-1}$
• so we finally have $\Big[ \ I \; \Big| \; A^{-1} \ \Big]$

Cramer's Rule

• We can compute the inverse of $A$ using the following formula:
• $A^{-1} = \cfrac{1}{| A |} C^T$
• where $|A|$ is the Determinant of $A$ and $C^T$ is the Cofactors matrix

Properties

• $(AB)^{-1} = B^{-1} A^{-1}$
• $(A^{-1})^T = (A^T)^{-1}$