Matrix Transformation Examples and Practice Problems

Find the production matrix for the following input-output and demand matrices using the open model.
$A=\left[\begin{array}{cc}-<!--\; -\; -->0.1& 0.2\\ 0.55& 0.4\end{array}\right]$
$D=\left[\begin{array}{c}3\\ 4\end{array}\right]$

Solve the following system of equations by using the inverse of the coefficient matrix A.
(AX=B), x+5y=-10, -2x+7y=-31

If A is an n x n matrix , where are the entries on the main diagonal of A-A^T? Justify yoyr answer.

Let $T:M_{2x2}\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^3$ have matrix $[T{]}_{B,A}=\left[\begin{array}{cccc}1& 2& 0& 1\\ 0& 1& 1& 0\\ 1& 1& -<!--\; -\; -->1& -<!--\; -\; -->1\end{array}\right]$ relative to $A=\{\left[\begin{array}{cc}2& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 5& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 6\end{array}\right]\}$ and $\mathrm{\beta <!--\; \beta \; -->}=\{(1,1,1),(1,2,3),(1,4,9)\}$. Find the matrix of T relative to the bases $A^{\prime }=\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 4\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 2& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 7\end{array}\right]\}$ and ${\mathrm{\beta <!--\; \beta \; -->}}^{\prime }=\{(1,1,1),(1,0,0),(1,1,0)\}$

representative matrix of a linear transformation
given a linear transformation: $T:M_n(\mathbb{C})\to <!--\; \to \; -->M_n(\mathbb{C})$, $T(A)=A-<!--\; -\; -->2A^T$, what is the

Let V be a 3 dimensional vector space over a field F and fix $({\mathbf{v}}_{\mathbf{1}},{\mathbf{v}}_{\mathbf{2}},{\mathbf{v}}_{\mathbf{3}})$ as a basis. Consider a linear transformation $T:V\to <!--\; \backslash rightarrow\; -->V$. Then we have
$T({\mathbf{v}}_{\mathbf{1}})=a_{11}{\mathbf{v}}_{\mathbf{1}}+a_{21}{\mathbf{v}}_{\mathbf{2}}+a_{31}{\mathbf{v}}_{\mathbf{3}}$
$T({\mathbf{v}}_{\mathbf{2}})=a_{12}{\mathbf{v}}_{\mathbf{1}}+a_{22}{\mathbf{v}}_{\mathbf{2}}+a_{32}{\mathbf{v}}_{\mathbf{3}}$
$T({\mathbf{v}}_{\mathbf{3}})=a_{13}{\mathbf{v}}_{\mathbf{1}}+a_{23}{\mathbf{v}}_{\mathbf{2}}+a_{33}{\mathbf{v}}_{\mathbf{3}}$
So that we can identify T by the matrix
$\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right)$
But then when I read several linear algebra book, it said if $T({\mathbf{v}}_{\mathbf{i}})=\underset{j}{\sum <!--\; \sum \; -->}{a}_{ij}{\mathbf{v}}_{\mathbf{j}}$ , then we can identify T by the matrix $(a_{ij})$. My problem is: isn't the matrix is $(a_{ji})$ instead of $(a_{ij})$?

Let A be the matrix below and define a transformation $T:{\mathbb{R}}^3\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^3$ by $T(U)=AU.$. For each of the vectors $B$ below, find a vector $U$ such that $T$ maps $U$ to $B$, if possible. Otherwise state that there is no such $U$.
$\left(\begin{array}{ccc}1& -<!--\; -\; -->3& 2\\ 2& -<!--\; -\; -->4& 4\\ 3& -<!--\; -\; -->8& 6\end{array}\right)=A$
a) $\left(\begin{array}{c}4\\ 6\\ 11\end{array}\right)=B$
b) $\left(\begin{array}{c}-<!--\; -\; -->3\\ -<!--\; -\; -->2\\ -<!--\; -\; -->7\end{array}\right)=B$

Suppose $T$ is a transformation from ${\mathbb{R}}^2$ to ${\mathbb{R}}^2$. Find the matrix $A$ that induces $T$ if $T$ is the (counter-clockwise) rotation by $\frac{3}{4}\mathrm{\pi <!--\; \pi \; -->}$.
how to begin to find a matrix that is 2x2 for this question.

Let $A$ be the set of all $n\times <!--\; \times \; -->n$ symmetric real matirix and $f\in <!--\; \in \; -->C(\mathbb{R},A)$. Then whether there is a $g\in <!--\; \in \; -->C(\mathbb{R},O(n))$ such that for all $t\in <!--\; \in \; -->\mathbb{R}$, $g(t{)}^{-<!--\; -\; -->1}f(t)g(t)$ is a diagonal matrix?

Can't figure out this transformation matrix
1) The positive z axis normalized as Vector(0,0,1) has to map to an arbitrary direction vector in the new coordinate system Vector(a,b,c)
2) The origin in the original coordinate system has to map to an arbitrary position P in the new coordinate system.
3) This might be redundant but the positive Y axis has to map to a specific direction vector(d,e,f) which is perpendicular to Vector(a,b,c) from before.
So my question is twofold: 1) How would I go about constructing this transformation matrix and 2) Is this enough data to ensure that any arbitrary vector in coordinate system 1 will be accurately transformed in coordinate system 2?

How can one prove the following identity of the cross product?
$(Ma)\times <!--\; \times \; -->(Mb)=det(M)(M^{\mathrm{T}}{)}^{-<!--\; -\; -->1}(a\times <!--\; \times \; -->b)$
$a$ and $b$ are 3-vectors, and $M$ is an invertible real $3\times <!--\; \times \; -->3$ matrix.

I'd like to be able to enter a vector or matrix, see it in 2-space or 3-space, enter a transformation vector or matrix, and see the result. For example, enter a 3x3 matrix, see the parallelepiped it represents, enter a rotation matrix, see the rotated parallelepiped.

Let $M$ be a transformation matrix $B\to <!--\; \backslash rightarrow\; -->B^{\prime }$
discovered that $M^{-<!--\; -\; -->1}$ is the opposite transformation.
What makes it true?

Consider the transformation $T:{\mathbb{R}}^3\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^2$ defined by:
$T(x)=T\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=(2x_1+x_3)\left(\begin{array}{c}1\\ 2\end{array}\right)+(x_2-<!--\; -\; -->3x_3)\left(\begin{array}{c}-<!--\; -\; -->1\\ 1\end{array}\right)$
1a) determine the matrix of the above transformation
1b) determine the reduced row echelon form of the matrix found in 1a
1c) based on your answer to part 1b, is the transformation T onto?
1d) based on your answer to part 1b, is the transformation T one-to-one?
1e) based on your answer to part 1b, determine the set of vectors $x$ in ${\mathbb{R}}^3$ for which $T(x)=0$. Write your answer in parametric vector form

Can't figure out this transformation matrix
1) The positive z axis normalized as Vector(0,0,1) has to map to an arbitrary direction vector in the new coordinate system Vector(a,b,c)
2) The origin in the original coordinate system has to map to an arbitrary position P in the new coordinate system.
3) This might be redundant but the positive Y axis has to map to a specific direction vector(d,e,f) which is perpendicular to Vector(a,b,c) from before.
So my question is twofold: 1) How would I go about constructing this transformation matrix and 2) Is this enough data to ensure that any arbitrary vector in coordinate system 1 will be accurately transformed in coordinate system 2?

Find the transformation matrix
Let $B=\{2x,3x+x^2,-<!--\; -\; -->1\},B^{\prime }=\{1,1+x,1+x+x^2\}$
Need to find the transformation matrix from $B$ to $B^{\prime }$.
I know that:
$(a{x}^2+bx+c{)}_B=(\frac{b-<!--\; -\; -->3c}{2},c,-<!--\; -\; -->a)$
$(a{x}^2+bx+c{)}_{B^{\prime }}=(a-<!--\; -\; -->b,b-<!--\; -\; -->c,c)$
How to proceed using this info in order to find the transformation matrix?

Let $\mathrm{\beta <!--\; \beta \; -->}$ and ${\mathrm{\beta <!--\; \beta \; -->}}^{\prime }$ be bases for the finite dimensional vector space $V$ of dimension n over the field $\mathbb{F}$, and let $Q=[I_V{]}_{{\mathrm{\beta <!--\; \beta \; -->}}^{\prime }}^{\mathrm{\beta <!--\; \beta \; -->}}$, where $I_V$ is the identity operator on $V$. I just recently proved that $Q[x{]}_{{\mathrm{\beta <!--\; \beta \; -->}}^{\prime }}=[x{]}_{\mathrm{\beta <!--\; \beta \; -->}}$ for every $x\in <!--\; \in \; -->V$ (twas rather simple), which suggests the title of "basis transformation" matrix or "coordinate transformation" matrix for the matrix $Q$. I am now wondering whether the converse holds.
Let $V^{\prime }$ denote $V$ with its elements written with respect to the basis ${\mathrm{\beta <!--\; \beta \; -->}}^{\prime }$, and suppose that $Q\in <!--\; \in \; -->M_n(\mathbb{F})$ is such that $Q[x{]}_{{\mathrm{\beta <!--\; \beta \; -->}}^{\prime }}=[x{]}_{\mathrm{\beta <!--\; \beta \; -->}}$ for every $x$. Since $\mathcal{L}(V^{\prime }V)$ is isomorphic to the matrix algebra $M_n(\mathbb{F})$ by sending a linear operator to its matrix representation, given $Q$ there exists a linear operator $T\in <!--\; \in \; -->\mathcal{L}(V^{\prime },V)$ such that $Q=[T{]}_{{\mathrm{\beta <!--\; \beta \; -->}}^{\prime }}^{\mathrm{\beta <!--\; \beta \; -->}}$. Hence $[T{]}_{{\mathrm{\beta <!--\; \beta \; -->}}^{\prime }}^{\mathrm{\beta <!--\; \beta \; -->}}[x{]}_{{\mathrm{\beta <!--\; \beta \; -->}}^{\prime }}=[x{]}_{\mathrm{\beta <!--\; \beta \; -->}}$ or $[T(x){]}_{\mathrm{\beta <!--\; \beta \; -->}}=[x{]}_{\mathrm{\beta <!--\; \beta \; -->}}$...
I want to say that this implies $T=I_V$, but I can't clearly see what lemma I need in order to make that conclusion.

Consider the 4-by-4 matrix $\mathit{M}={\mathit{M}}_0+{\mathit{M}}_1$, where
${\mathit{M}}_0=\mathrm{\alpha <!--\; \alpha \; -->}\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& -<!--\; -\; -->1& 0\\ 0& 0& 0& -<!--\; -\; -->1\end{array}\right)$ and ${\mathit{M}}_1=\mathrm{\beta <!--\; \beta \; -->}\left(\begin{array}{cccc}0& \mathrm{\gamma <!--\; \gamma \; -->}& 0& -<!--\; -\; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^{\ast <!--\; \ast \; -->}\\ {\mathrm{\gamma <!--\; \gamma \; -->}}^{\ast <!--\; \ast \; -->}& 0& -<!--\; -\; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^{\ast <!--\; \ast \; -->}& 0\\ 0& \mathrm{\gamma <!--\; \gamma \; -->}& 0& -<!--\; -\; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^{\ast <!--\; \ast \; -->}\\ \mathrm{\gamma <!--\; \gamma \; -->}& 0& -<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}& 0\end{array}\right)$
where $\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}$ are constants and $\mathrm{\gamma <!--\; \gamma \; -->}={\mathrm{\gamma <!--\; \gamma \; -->}}_x+i{\mathrm{\gamma <!--\; \gamma \; -->}}_y$ is complex.
Is it possible to unitary transform $\mathit{M}$ into block off-diagonal form ${\mathit{M}}_B$?
Namely, I want to find a unitary transform $\mathit{U}$ so that I can write down ${\mathit{M}}_B=\mathit{U}\mathit{M}{\mathit{U}}^{\ast <!--\; \ast \; -->}$ (here ${\mathit{U}}^{\ast <!--\; \ast \; -->}$ is the conjugate transpose).
Explicitly, the required block off-diagonal matrix is (in general form)
${\mathit{M}}_B=\left(\begin{array}{cc}0& \mathit{Q}\\ {\mathit{Q}}^{\ast <!--\; \ast \; -->}& 0\end{array}\right)$ where $\mathit{Q}=\left(\begin{array}{cc}{Q}_z& Q_x-<!--\; -\; -->i{Q}_y\\ Q_x+i{Q}_y& -<!--\; -\; -->Q_z\end{array}\right)$
Is there a general recipe to find such a unitary transformation matrix $\mathit{U}$ which leads to the block off-diagonal form, $\mathit{M}\to <!--\; \backslash rightarrow\; -->{\mathit{M}}_B$?

The transformation $\mathbf{\text{T}}$ maps points $(x,y)$ of the plane into image points $(x^{\prime },y^{\prime })$ such that
$\begin{array}{rl}{x}^{\prime }& =4x+2y+14\\ y^{\prime }& =2x+7y+42\end{array}$
Find the coordinates of the invariant point of $\mathbf{\text{T}}$. Hence express $\mathbf{\text{T}}$ in the form
$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }+k\end{array}\right)=\mathbf{\text{A}}\left(\begin{array}{c}x\\ y+k\end{array}\right)$
where $k$ is a positive integer and $\mathbf{\text{A}}$ is a $2\times <!--\; \times \; -->2$ matrix.