Determine whether the statement is true or false.If the last row of the reduced row echelon form of an augmented matrix of a system of linear equation

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

The coefficient matrix for a system of linear differential equations of the form ${\displaystyle y^1=A_y}$ has the given eigenvalues and eigenspace bases. Find the general solution for the system.

The reduced row echelon form of the augmented matrix of a system of linear equations is given. Tell whether the system has one solution, no solution, or infinitely many solutions. Write the solutions or, if there is no solution, say the system is inconsistent. $\left[\begin{array}{ccccc}1& 0& -2& |& 6\\ 0& 1& 3& |& 1\end{array}\right]$

Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given reduced row echelon form. Solve the system. Assume that the variables are named $x_1,x_2,...x,...$ ,… from left to right. 

Determine whether the statement is true or false. If the last row of the reduced row echelon form of an augmented matrix of a system of linear equations has only zero entries, then the system has infinitely many solutions.

Write the vector form of the general solution of the given system of linear equations.
x1+2x2−x3+x4=0