This has four parts to it. the following system oflinear equations x - y =2z=13 2x +2y -z =-6 -x + 3y +z = -7 a) Provide a coefficient matrix corresponding to the system oflinear equations b) what is the inverse of this matrix? c) What is the transpose of the matrix? d) find the determinant for this matrix?

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

Find the area of the parallelogram with vertices A(-3, 0), B(-1 , 3), C(5, 2), and D(3, -1).

Solve the system
${\displaystyle 3x_1+2x_2-x_3=4}$
${\displaystyle x_1-2x_2+2x_3=1}$
${\displaystyle 11x_1+2x_2+x_3=14}$

On the one hand, we always use the multiplication by transformation matrix when we want to receive a matrix in a new basis:
- $A^{\prime }=CA$ , where $A^{\prime }$ is a matrix in a new basis; $C$ is a transformation matrix; $A$ is a matrix in the standard basis.
However, when we speak about getting the matrix in its eigenbasis, we use the other formula.
- $A=C^{-1}{A}^{\prime }C$ where $A$ is a matrix in a standard basis; $A^{\prime }$ is a matrix in the eigenbasis; $C$ is the transformation matrix.
Why do we have different formulas while doing the same thing (matrix transformations) or do I get something wrong?

Suppose $T$ is an element of $L(P_3(\mathbb{R}),P_2(\mathbb{R}))$ is the differentiation map defined by $Tp=p^{\prime }$. Find a basis of $P_3(\mathbb{R})$ and a basis of $P_2(\mathbb{R})$ such that the matrix of T with respect to these basis is
$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\end{array}\right)$

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 $x_1,x_2,\dots$from left to right. $\left[\begin{array}{ccccc}1& 0& 5& 3& 2\\ 0& 1& -2& 4& -7\\ 0& 0& 1& 1& 3\end{array}\right]$

Linear transformation and its matrix
two bases:
$A=\{v_1,v_2,v_3\}$ and $B=\{2v_1,v_2+v_3,-v_1+2v_2-v_3\}$
There is also a linear transformation: $T:{\mathbb{R}}^3\to {\mathbb{R}}^3$
Matrix in base $A$:
$M_T^A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 1& 1& 0\end{array}\right]$
Now I am to find matrix of the linear transformation $T$ in base $B$.
I have found two transition matrixes (from base $A$ to $B$ and from $B$ to $A$):
$P_A^B=\left[\begin{array}{ccc}2& 0& -1\\ 0& 1& 2\\ 0& 1& -1\end{array}\right]$
$(P_A^B{)}^{-1}=P_B^A=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{6}& \frac{-1}{6}\\ 0& \frac{1}{3}& \frac{2}{3}\\ 0& \frac{1}{3}& \frac{-1}{3}\end{array}\right]$
How can I find $M_T^B$?