Use an inverse matrix to solve system of linear equations. (a) 2x-y=-3 2x+y=7 (b) 2x-y=-1 2x+y=-3

Find the linear approximation of the function $f(x)=\sqrt{4-x}$ at $a=0$
Use L(x) to approximate the numbers $\sqrt{3.9}$ and $\sqrt{3.99}$ Round to four decimal places

I know from linear algebra that for a non-homogeneous consistent linear system that has an infinite number of solutions, if the last row of the augmented matrix has all zeros, then we have an infinite number of particular solutions.
I was wondering if there is a similar parallel with differential equations, namely say I have a differential equation of the form
$y^{″}(x)(Ax+B{)}^2+y(x)=$ some cubic, and I want to find a particular solution of this. Then I make an $ansatz$ that my $y(x)$ is some cubic as well, of the form $y(x)=a_{0}x+a_{1}x+...a_3{x}^3$
When comparing coeffcients on the left and right hand sides, I have come across cases where by the coeffcients $a_0$, ... $a_1$ on the L.H.S. do not have particular values, but are instead expressed as expressions involving one of the other coefficients as a free variable.
Is it possible that differential equations too may have an infinite number of particular solutions?

I am having trouble understanding the relatioship between rows and columns of a matrix.
Say, the following homogeneous system has a nontrivial solution.
$$\begin{gathered} 3x_1+5x_2-4x_3=0 \\ -3x_1-2x_2+4x_3=0 \\ 6x_1+x_2-8x_3=0 \end{gathered}$$
Let A be the coefficient matrix and row reduce [A0] to row-echelon form:
$\left[\begin{array}{cccc}3& 5& -4& 0\\ -3& -2& 4& 0\\ 6& 1& -8& 0\end{array}\right]\to \left[\begin{array}{cccc}3& 5& -4& 0\\ 0& 3& 0& 0\\ 0& 0& 0& 0\end{array}\right]$
$a1a2a3$
Here, we see $x_3$ is a free variable and thus we can say 3rd column,$a_3$, is in $\text{span}(a_1,a_2)$
But what does it mean for an echelon form of a matrix to have a row of 0's?
Does that mean 3rd row can be generated by 1st & 2nd rows?
just like 3rd column can be generated by 1st & 2nd columns?
And this raises another question for me, why do we mostly focus on columns of a matrix?
because I get the impression that ,for vectors and other concepts, our only concern is
whether the columns span ${\mathbb{R}}^n$ or the columns are linearly independent and so on.
I thought linear algebra is all about solving a system of linear equations,
and linear equations are rows of a matrix, thus i think it'd be logical to focus more on rows than columns. But why?

To determine whether a system of linear equations lacks a solution, we can convert it into reduced row echelon form. If the bottom-right value is 1, then the system lacks solutions.
Why is this true? It just seems like a random indicator to me.

Determine if (1,3) is a solution to the given system of linear equations.
${\displaystyle 5x+y=8}$
${\displaystyle x+2y=5}$

Theorem:
Given a system of linear equations ${\displaystyle Ax=b}$ where
${\displaystyle A\in M_{m\times n}\left(\mathbb{R}\right),x\in {\mathbb{R}}_{\text{col}}^n,b\in {\mathbb{R}}_{\text{col}}^m}$.
Deduce that a solution x exists if and only if ${\displaystyle \text{rank}\left(A\mid b\right)=\text{rank}\left(A\right)}$ where ${\displaystyle A\mid b}$ is the augmented coefficient matrix of this system
I am having trouble proving the above theorem from my Linear Algebra course, I understand that A|b must reduce under elementary row operations to a form which is consistent but I don't understand exactly why the matrix A|b need have the same rank as A for this to happen.
Please correct me if I am mistaken

Solve the following system of linear equations in terms of parameter $a\in \mathbb{R}$ and explain geometric interpretation of this system:
$ax+y+z=1,2x+2ay+2z=3,x+y+az=1$.
By Cronecker Capelli's theorem, we get:
$\left[\begin{array}{cccc}a& 1& 1& 1\\ 2& 2a& 2& 3\\ 1& 1& a& 1\end{array}\right]$
Row echelon form of this matrix is
$\left[\begin{array}{cccc}1& 1& a& 1\\ 0& 2(a-1)& 2(1-a)& 1\\ 0& 0& (1-a)(a+2)& (3-2a)/2\end{array}\right]$
System is inconsistent for $a=1\vee a=-2$. For every other value of $a$, system has unique solution.
For every $a$ instead of $a=1\wedge a=-2$ there are three planes that intersect at a point.
Question: Is this geometric interpretation correct?