Using inverses to solve systems of equations

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this,

Can someone please tell me where they got $X = IX$ from?

Many thanks!

 

[fresh_42]

From the definition of $I$. It is the one in matrix multiplication.
$$
\begin{pmatrix}x_{11}&\ldots &x_{1n}\\ \vdots&&\vdots \\x_{n1}&\ldots &x_{nn}\end{pmatrix} \cdot \underbrace{\begin{pmatrix}1&0&\ldots&0&0\\ 0&1&\ldots&0&0 \\ \vdots&\vdots&\ddots&\vdots&\vdots \\
0&0&\ldots&0&1 \end{pmatrix}}_{=I}=\begin{pmatrix}x_{11}&\ldots &x_{1n}\\ \vdots&&\vdots \\x_{n1}&\ldots &x_{nn}\end{pmatrix}
$$

Then they used $A^{-1} A= I$ and the associativity law: $(I\cdot X)=((A^{-1}A)X)=(A^{-1}(AX))=A^{-1}B.$

 

[ChiralSuperfields]

From the definition of $I$. It is the one in matrix multiplication.
$$
\begin{pmatrix}x_{11}&\ldots &x_{1n}\\ \vdots&&\vdots \\x_{n1}&\ldots &x_{nn}\end{pmatrix} \cdot \underbrace{\begin{pmatrix}1&0&\ldots&0&0\\ 0&1&\ldots&0&0 \\ \vdots&\vdots&\ddots&\vdots&\vdots \\
0&0&\ldots&0&1 \end{pmatrix}}_{=I}=\begin{pmatrix}x_{11}&\ldots &x_{1n}\\ \vdots&&\vdots \\x_{n1}&\ldots &x_{nn}\end{pmatrix}
$$

Then they used $A^{-1} A= I$ and the associativity law: $(I\cdot X)=((A^{-1}A)X)=(A^{-1}(AX))=A^{-1}B.$

Thank you for your help @fresh_42! I see now.

 

[Mark44]

Can someone please tell me where they got $X = IX$ from?

This should be obvious. Based on the initial post, X is a column vector. If X consists of n elements, multiplication of X by an n x n identity matrix $I_n$ produces exactly the same vector X. This is analogous to writing $b = 1 \cdot b$ for ordinary numbers.

 

[YouAreAwesome]

Homework Statement: Please see below
Relevant Equations: Please see below

For this,
View attachment 325404
Can someone please tell me where they got $X = IX$ from?

Many thanks!

The $I$ should trigger the phrase "$I$dentity Matrix". It was the reason $I$ was chosen rather than some other letter in the first place.