MATH 110 - Day 12

2024-10-04 23:29:13 -0400 EDT

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Inverse Matrixes

Let A be an $ n \times n$ matrix

The inverse of A is denoted by $A^{-1}$

if $AB = I_n =BA$, then $B = A^{-1}$ and A is invertible

Note: This only works with square matrixes

Also Note: there is only one inverse of a matrix (if it is invertible)

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An application:
consider a system of n linear equations with m variables
$ \circledast = \left[ \begin{array}{rrrr} a_{11} & a_{12} & \dots & a_{1n} \\ a_{11} & a_{12} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{array} \right] $

Let A be a $n \times n$ matrix, $\vec{x} be a $n \times 1$ matrix, $\vec{b}$ be a $1 \times m$ matrix
then $\circledast$ could be written as $A \vec{x} = \vec{b}$

Suppose that A is invertible ($A^{-1}$ exist). Multiply $A^{-1}$ on both sides
$A^{-1}A \vec{x} = A^{-1} \vec{b} \Rightarrow \vec{x} = A^{-1} \vec{b}$

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Question: How can you find the inverse of A if A is invertible?

Let $ A = \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right], B = \left[ \begin{array}{rr} x & y \\ z & w \end{array} \right], AB = BA = I_n$

$ \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] \left[ \begin{array}{rr} x & y \\ z & w \end{array} \right] = \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right] $

$\Rightarrow$

$ax + bz = 1$
$ay + bw = 0$
$cx + dz = 0$
$cy + dw = 1$

$\Rightarrow$

$ \left[ \begin{array}{rrrrr} a & 0 & b & 0 & | & 1 \\ 0 & a & 0 & b & | & 0 \\ c & 0 & d & 0 & | & 0 \\ 0 & c & 0 & d & | & 1 \end{array} \right] \Rightarrow A^{-1} = \frac{1}{ad - bc} \left[ \begin{array}{rrrrr} d & -b \\ -c & a \end{array} \right] $

Example:
2x + y = -1
5x + 3y = 2

$ \left[ \begin{array}{rrr} 2 & 1 & | & -1 \\ 5 & 3 & | & 2 \end{array} \right] $

$ \left[ \begin{array}{rrr} 2 & 1 \\ 5 & 3 \end{array} \right]^{-1} = \left[ \begin{array}{rrr} 3 & -1 \\ -5 & 2 \end{array} \right] \Rightarrow \vec{x} = \left[ \begin{array}{rrr} 3 & -1 \\ 5 & 2 \end{array} \right] \left[ \begin{array}{rrr} -1 \\ 2 \end{array} \right] = \left[ \begin{array}{rrr} -5 \\ 9 \end{array} \right] $

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Algebra for Inverse Matrixes

1. $(A^{-1})^{-1} = A$
2. $(cA)^{-1} = \frac{1}{c} A^{-1}$
3. $(AB)^{-1} = B^{-1} A^{-1}$
4. $(A^T)^{-1} = (A^{-1})^T$
5. $(A^n)^{-1} = (A^{-1})^n$

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The Fundamental Theorem of Invertible Matrixes

if $A$ i a square matrix of $n$ size and $I$ is the identity matrix of $n$ size, then:

1. A is invertible
2. $A \vec{x} = \vec{b}$ is a unique solution for $\vec{b} \in \mathbb{R}^n$
3. $A \vec{x} = \vec{o}$ has only one trivial solution
4. RREF of $A$ is $I$
5. A is a product of Elementary Matrices

Algorithm for Finding the Inverse of a Matrix

Algo:
input: Matrix of size $n$ output: If the matrix is invertible

Step 1: Form a $ n \times 2n$ matrix

$ [ A | I ] $

Step 2: Perform row operations to turn $A$ to RREF

$ [ A | I ] \Rightarrow [ E | B ]$

if $E = I_n$, then B = A^{-1}

Example:
$A = \left[ \begin{array}{rrr} 1 & -1 & 2 \\ 3 & 1 & 2 \\ 2 & 3 & -2 \end{array} \right] $

Step 1: Form a $ n \times 2n$ matrix
$ A = \left[ \begin{array}{rrr} 1 & -1 & 2 & | & 1 & 0 & 0 \\ 3 & 1 & 2 & | & 0 & 1 & 0 \\ 2 & 3 & -2 & | & 0 & 0 & 1 \end{array} \right] $

Step 2: Perform row operations to turn $A$ to RREF $ \left[ \begin{array}{rrr} 1 & -1 & 2 & | & 1 & 0 & 0 \\ 0 & 1 & -1 & | & - \frac{3}{4} & \frac{1}{4} 0 \\ 0 & 0 & 0 & | & \frac{7}{20} & - \frac{1}{4} & \frac{1}{5} \end{array} \right] \neq I_n $

Step 3: $\therefore A$ is not Invertible

You can verify your matrix is correct with $AA^{-1} = I_n$