New concepts:

• Gauss-Jordan algorithm

• Existence of matrix inverse

• Operations with inverses

Recall that the Gaussian multiplier matrix ...

... has inverse (matrix that “undoes” the linear transformation)

What about general square matrices $𝑨\in {ℝ}^{m\times m}$? How to find inverse

$𝑿$ is inverse if $𝑨𝑿=𝑰$ or

Find the inverse is equivalent to solving systems $𝑨{𝒙}_1={𝒆}_1$, ...,$𝑨{𝒙}_m={𝒆}_m$

Gauss Jordan algoritm generalizes Gaussian elimination that solves a single linear system to solving $m$ systems simultaneously by forming the bordered matrix $\left[\begin{array}{lll}𝑨& |.& 𝑰\end{array}\right]$

Example

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A=[1 2 1; -1 0 2; 2 -1 -4]; AX=[A eye(3)]; format rat; disp(AX);

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When does a matrix inverse exist? $𝑨\in {ℝ}^{m\times m}$

1. $𝑨$ invertible

2. $𝑨𝒙=𝒃$ has a unique solution for all $𝒃\in {ℝ}^m$

3. $𝑨𝒙=𝟎$ has a unique solution

4. The reduced row echelon form of $𝑨$ is $𝑰$

5. $𝑨$ can be written as product of elementary matrices

$a\Rightarrow b$

$𝑨$ invertible $\Rightarrow$ ${𝑨}^{-1}$ exists, and $𝒙={𝑨}^{-1}𝒃$ is a solution $𝑨({𝑨}^{-1}𝒃)=(𝑨{𝑨}^{-1})𝒃=𝒃$. If there were two solutions $𝒙,𝒚$, then

${\displaystyle 𝒙-𝒚=\left({𝑨}^{-1}𝑨\right)(𝒙-𝒚)={𝑨}^{-1}(𝑨𝒙-𝑨𝒚)={𝑨}^{-1}(𝒃-𝒃)={𝑨}^{-1}𝟎=𝟎.}$

$b\Rightarrow c$

Choose $𝒃=𝟎$

$c\Rightarrow d$

$\left[\begin{array}{lll}𝑨& |.& 𝟎\end{array}\right]\sim \left[\begin{array}{lll}𝑼& |.& 𝟎\end{array}\right]$. If $𝑼\ne 𝑰$ there is a row of zeros, and solution is not unique. If solution is unique then $𝑼=𝑰$

$d\Rightarrow e$

$\left[\begin{array}{lll}𝑨& |.& 𝟎\end{array}\right]\sim \left[\begin{array}{lll}𝑰& |.& 𝟎\end{array}\right]$ implies ${𝑬}_k\dots {𝑬}_1𝑨=𝑰\Rightarrow 𝑨={𝑬}_1^{-1}\dots {𝑬}_k^{-1}$

$e\Rightarrow a$

$𝑨=$${𝑬}_1^{-1}\dots {𝑬}_k^{-1}\Rightarrow {𝑨}^{-1}={𝑬}_k\dots {𝑬}_1$.

The inverse of a product ${(𝑨𝑩)}^{-1}={𝑩}^{-1}{𝑨}^{-1}$

If $𝑨\in {ℝ}^{m\times m}$ invertible so are: $c𝑨$, ${𝑨}^T$, ${𝑨}^k$

Verify