The reduced row echelon form of the augmented matrix of a system of linear equations is given. Determine whether this system of linear equations is co

Determine whether the given set S is a subspace of the vector space V.
A. V=$P_5$, and S is the subset of $P_5$ consisting of those polynomials satisfying p(1)>p(0).
B. $V=R_3$, and S is the set of vectors $(x_1,x_2,x_3)$ in V satisfying $x_1-6x_2+x_3=5$.
C. $V=R^n$, and S is the set of solutions to the homogeneous linear system Ax=0 where A is a fixed m×n matrix.
D. V=$C^2(I)$, and S is the subset of V consisting of those functions satisfying the differential equation y″−4y′+3y=0.
E. V is the vector space of all real-valued functions defined on the interval [a,b], and S is the subset of V consisting of those functions satisfying f(a)=5.
F. V=$P_n$, and S is the subset of $P_n$ consisting of those polynomials satisfying p(0)=0.
G. $V=M_n(R)$, and S is the subset of all symmetric matrices

Each of the matrices is the final matrix form for a system of two linear equations in the variables $x_1$ and $x_2$. Write the solution of the system.

$\left[\begin{array}{cccc}1& 3& |& 2\\ 0& 0& |& 4\end{array}\right]$

At school, or in a first-year course on DEs, we learn (perhaps in less abstract language) that if you have a linear $n$th-order differential equation
$Ly=f$
then the general solution is something of the form
$y=a_1{y}_1+...+a_n{y}_n+g$
where the $y_i$ are independent and satisfy $L{y}_i=0$, and $g$ satisfies $Lg=f$. Then we receive lots of training in how to find the $y_i$ and $g$.
Obviously any choice of the $a_i$ will give us a solution to $Ly=f$. But how do you know that there aren't any more solutions?
We justify this by making an analogy with systems of linear equations $Ax=b$, saying something along the lines of 'the space of solutions has the same dimension as the kernel of $A$'. But that works in finite dimensions - how do we know that the same is true with linear operators?

Solve the system by clennaton

$\{\begin{array}{l}2x-y=0\\ 3x-2y=-3\end{array}$
The solution is____

I have a non-linear form of the Poisson equation (with a diffusion coefficient that is a function of the derivatives of the dependent variable) that I'm trying to solve numerically (using FEM which requires the pde to be posed in its weak form).
$\mathrm{\nabla }\cdot (\eta \mathrm{\nabla }v)=G(x)$
where $\eta =\sqrt{1/2(\mathrm{\nabla }v:\mathrm{\nabla }v)}$ and $v=v(y,z)$. Multiplying by a test function θ and integrating wrt the domain, $\mathrm{\Omega }$, gives the weak form
${\oint }_{\mathrm{\Gamma }}\theta \cdot (\eta \mathrm{\nabla }v\cdot \mathbf{n})d\mathrm{\Gamma }-{\int }_{\mathrm{\Omega }}(\eta \mathrm{\nabla }v\cdot \mathrm{\nabla }\theta +G)d\mathrm{\Omega }=0$
However, none of my boundary conditions correspond to the Dirichelet type as I have ${\mathrm{\partial }}_{n}v=v$ on three of the boundaries in a rectangular domain and ${\mathrm{\partial }}_{n}v=0$ on the other (n is the normal vector to surface). Is this problem actually solvable using FEM, or at all? I have the solution to a simpler problem which might be a reasonable guess at the solution here.

Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given row echelon form. Solve the system by back substitution. Assume that the variables are named x1,x2,$\dots$ from left to right.
$\left[\begin{array}{cccc}1& -3& 7& 1\\ 0& 1& 4& 0\\ 0& 0& 0& 1\end{array}\right]$

The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution.
${\displaystyle \left[\begin{array}{cccc}1& 0& -1& -2\\ 0& 1& 2& 3\end{array}\right]}$