Matrix Inverses

Def :drill: Scheduled: 2022-10-22 A is a square matrix. A matrix B is called an [inverse] if AB=I and BA=I (where I is the identity matrix). If it is.. then it can be called an [invertable] matrix

ex. \begin{bmatrix} -1&1\ 0&1 \end{bmatrix} is the inverse of \begin{bmatrix} 0&1\ 1&1 \end{bmatrix}

Theorem :drill: If A is invertible, then the inverse is [unique].

notation

The inverse of A is $A^{-1}$ which means $A{A}^{-1}=I$ and $A^{-1}A=I$.

If A= \begin{bmatrix} a&b\c&d \end{bmatrix}, define det A = det \begin{bmatrix} a&b\c&d \end{bmatrix}

= ad-be

Adjugate of A = \begin{bmatrix} d&-b\-c&a \end{bmatrix}

Theorem :drill:

\begin{math} A=\frac{1}{\text{det} A}\text{adj} A \end{math}

as long as [det A != 0]

EX

Find A^-1 if A = \begin{bmatrix} 2&4\ -3&8 \end{bmatrix}

det A = 16 + 12 = 28 != 0

A^-1 = 1/det A adj A \begin{math} = 1/28 \begin{bmatrix} 8&-4\3&2 \end{bmatrix}

\begin{bmatrix} 8/28&-4/28\ 3/28&2/28\ \end{bmatrix} \end{math}

Inverses and linear systems

Suppose the system Ax=b where A and b are known and x is determined. If A is invertable.

\begin{math} Ax=b\ A^{-1}Ax=A^{-1}b\ Ix=A^{-1}b\ x=A^{-1}b \end{math}

Theorem :drill: Suppose a system in N variables is written in matrix form as Ax=b. If the n*n coefficient matrix A is invertible, the system has a unique solution given by [$x=A^{-1}b$]

Inversion method

If A is an [invertible matrix], there exists a sequence of elementary row operations that carry matrix A to the identity matrix I.

\begin{align*} A \rightarrow I \end{align*} This same series of row operations carries I to $A^{-1}$, that is $I\to A^{-1}$.

In summary, $[AI]\to [I{A}^{-1}]$

Example

Use the inverse algorithm to find the inverse of the matrix given by:

\begin{align*} A= \begin{bmatrix} 2&7&1\ 1&4&-1\ 1&3&0 \end{bmatrix} \end{align*}

—

Apply elementary row operations to the double matrix.

\begin{bmatrix} 2&7&1&|&1&0&0\ 1&4&-1&|&0&1&0\ 1&3&0&|&0&0&1\ \end{bmatrix}

r2 ←> r1 \begin{bmatrix} 1&4&-1&|&0&1&0\ 2&7&1&|&1&0&0\ 1&3&0&|&0&0&1\ \end{bmatrix}

r2 - 2*r1 r3 - r1 \begin{bmatrix} 1&4&-1&|&0&1&0\ 0&-1&3&|&1&-2&0\ 0&-1&1&|&0&-1&1\ \end{bmatrix}

r2 * -1 r3 = r3+r2 \begin{bmatrix} 1&4&-1&|&0&1&0\ 0&1&-3&|&-1&2&0\ 0&0&-2&|&-1&1&1\ \end{bmatrix}

r1 = r1-4*r2 r3 = r3/-2 \begin{bmatrix} 1&0&11&|&-4&7&0\ 0&1&-3&|&-1&2&0\ 0&0&1&|&1/2&-1/2&-1/2\ \end{bmatrix}

— i fucked this one up somethow.. r2 = r2 + 3 r3 r1 = r1 - 11r3 \begin{bmatrix} 1&0&0&|&-9.5&-3/2&11/2\ 0&1&0&|&1/2&1/2&3/2\ 0&0&1&|&1/2&-1/2&-1/2\ \end{bmatrix}

— copying from teacher \begin{bmatrix} 1&0&0&|&-3/2&-3/2&11/2\ 0&1&0&|&-1/2&1/2&-3/2\ 0&0&1&|&1/2&-1/2&-1/2\ \end{bmatrix}

Therefore the inverse matrix is:

\begin{align*} A^{-1} = \frac{1}{2} \begin{bmatrix} -3&-3&11\ -1&1&-3\ 1&-1&-1\ \end{bmatrix} \end{align*}

Go figure out how I messed up this step.

• TODO

Properties of inverses (theorem) :drill: Scheduled: 2022-10-22

1. $I$ is invertible and [$I^{-1}=I$|what is equal to I?]
2. If $A$ is invertible, so is it’s inverse $A^{-1}$ and $(A^{-1}{)}^{-1}=A$
3. If A and B are invertible, so is AB and the $(AB{)}^{-1}=B^{-1}{A}^{-1}$. Note, $(AB{)}^{-1}\ne A^{-1}{B}^{-1}$
4. If $A_1,A_2,...,A_k$ are all invertible, so is their [product $A_1{A}_2...A_k$. $(A_1{A}_2...A_k{)}^{-1}=A_k^{-1}...A^{-1}\ast 2A^{-1}\ast 1$.|operation]
5. If A is invertible, so is [A^k for any k >= 1 and $(A^k{)}^{-1}=(A^{-1}{)}^k$|operation].
6. If A is invertible and k is a != 0 is a number, then aA is invertible and [$(aA{)}^{-1}=\frac{1}{a}{A}^{-1}$ |notation].
7. If A is invertible, so is it’s transpose and $(A^T{)}^{-1}=(A^{-1}{)}^T$

Set up an org-drill for this.

• DONE Completed: 2022-10-11

Inverse Theorem

The following conditions are equivalent for a n*n matrix:

1. A is invertible
2. A homogeneous system Ax=0 has only a trivial solution x=0.
3. A can be carried to the identity matrix $I_n$ by elementary row operations.
4. The system Ax=b has at least one solution x for any choice of column b.
5. Since A is invertible, there exists an n*n matrix c such that $Ac=I$ (e.g. c is it’s inverse)

Proof of #2 \begin{align*} x= \begin{bmatrix} x_1\ x_2\ x_3\ \end{bmatrix} Ax=0\ A^{-1}Ax=A^{-1}0\ A^{-1}A = I\ Ix=A^{-1}0\ \text{Anything times zero is zero}\ x=0\ \end{align*}

Corollary :drill: An n*n matrix A is invertible iff [rank A=n|some property of the matrix]

Theorem

Let $T:{\mathbb{R}}^n\to {\mathbb{R}}^n$ denote the matrix transformation induced by an n*n matrix A. Then A is invertible iff T has an inverse. In this case, T has exactly one inverse, $T^{-1}$ and $T^{-1}:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is the transformation induced by the matrix $A^{-1}$. In other words, $(T_A{)}^{-1}=T_{(A^{-1})}$

how to show this

If A is an m*n matrix..

$T_A:{\mathbb{R}}^n\to {\mathbb{R}}^m$

$X->T_A(x)=Ax$ A is mn and X = n1. Resulting matrix is Mx1

TIL: Composition operator :drill: \begin{align*} f = A \rightarrow B\ g = B \rightarrow C\ f \circ g = A \rightarrow C\ (g \circ f)(x) = g(f(x))\ \end{align*}

$\circ$ is read as [“composed of”]

Ex: Use the inversion method to find A^-1

\begin{align*} A = \begin{bmatrix} 3&4&1\ 2&3&0\ 4&3&-1 \end{bmatrix} \end{align*}

Step 1: Convert to the identity matrix

\begin{bmatrix} 3&4&1&|&1&0&0\ 2&3&0&|&0&1&0\ 4&3&-1&|&0&0&1\ \end{bmatrix}

r1 /= 3 \begin{bmatrix} 1&4/3&1/3&|&1/3&0&0\ 2&3&0&|&0&1&0\ 4&3&-1&|&0&0&1\ \end{bmatrix}

r2 = r2-r12 r3 = r3-r14 \begin{bmatrix} 1&4/3&1/3&|&1/3&0&0\ 0&1/3&-2/3&|&-2/3&1&0\ 0&-7/3&-7/3&|&-4/3&0&1\ \end{bmatrix}

r2 *= 3 r3 *= 3 \begin{bmatrix} 1&4/3&1/3&|&1/3&0&0\ 0&1&-2&|&-2&3&0\ 0&-7&-7&|&-4&0&3\ \end{bmatrix}

r3 = r3 + 7*r2 \begin{bmatrix} 1&4/3&1/3&|&1/3&0&0\ 0&1&-2&|&-2&3&0\ 0&0&-21&|&-18&21&3\ \end{bmatrix}

r3 / -21 \begin{bmatrix} 1&4/3&1/3&|&1/3&0&0\ 0&1&-2&|&-2&3&0\ 0&0&1&|&18/21&1&-3/21\ \end{bmatrix}

r1 = r1 - r2*4/3 \begin{bmatrix} 1&0&3&|&3&-4&0\ 0&1&-2&|&-2&3&0\ 0&0&1&|&18/21&1&-3/21\ \end{bmatrix}

r2 = r2 + 2r3 r1 = r1 - 3r3 \begin{bmatrix} 1&0&0&|&9/21&-7&3/21\ 0&1&0&|&-6/21&5&-6/21\ 0&0&1&|&18/21&1&-3/21\ \end{bmatrix}

\begin{align*} = 1/21 \begin{bmatrix} 9& -7/21 & 3\ -6 & 5/21 & -6\ 18 & 1/21 & -3\ \end{bmatrix} \end{align*}