Semaj Christian
2022-06-25
Answered

$F:{P}_{2}\text{to}{P}_{2}$ to $F:{P}_{2}\text{to}{P}_{2}$ with the conditions that $F({p}_{0})=f$, $F({p}_{1})=g$, $F({p}_{2})=h$ where $f(x)={x}^{2}+3$, $g(x)={x}^{2}-x$, $h(x)=2+x$. Find the transformation matrix for $F$ in the basis $({p}_{0},{p}_{1},{p}_{2})$

You can still ask an expert for help

asked 2021-09-13

Assume that A is row equivalent to B. Find bases for Nul A and Col A.

asked 2021-09-18

Find an explicit description of Nul A by listing vectors that span the null space.

asked 2021-06-13

For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A.

$A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]$

Find a nonzero vector in Nul A.

$A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]$

Find a nonzero vector in Nul A.

asked 2022-05-12

In the vectorspace ${\mathbb{R}}^{2}$ two vectors are given (Which also form the basis a): ${a}_{1}=(8,-3)$ and ${a}_{2}=(5,-2)$

A linear mapping is determined by: $f({a}_{1})=2{a}_{1}-4{a}_{2}$ and $f({a}_{2})=-{a}_{1}+2{a}_{2}$

How can one determine the transformation matrix of f with respect to the standard e-basis?

A linear mapping is determined by: $f({a}_{1})=2{a}_{1}-4{a}_{2}$ and $f({a}_{2})=-{a}_{1}+2{a}_{2}$

How can one determine the transformation matrix of f with respect to the standard e-basis?

asked 2021-02-21

(7) If A and B are a square matrix of the same order. Prove that

asked 2022-08-02

Perform the indicated matrix operation.

$\left[\begin{array}{ccc}2x+y& x-2y& 4x\\ 3x& 3y& x+y\end{array}\right]+\left[\begin{array}{ccc}5x& 8y& 3x+y\\ 5x+5y& x& 2x\end{array}\right]$

$\left[\begin{array}{ccc}2x+y& x-2y& 4x\\ 3x& 3y& x+y\end{array}\right]+\left[\begin{array}{ccc}5x& 8y& 3x+y\\ 5x+5y& x& 2x\end{array}\right]$

asked 2022-07-05

$T({e}_{1})=T(1,0)=(\mathrm{cos}\theta ,\mathrm{sin}\theta )$

and

$T({e}_{2})=T(0,1)=(-\mathrm{sin}\theta ,\mathrm{cos}\theta )$

and

$A=[T({e}_{1})|T({e}_{2})]=\left[\begin{array}{cc}\mathrm{cos}\theta & -\mathrm{sin}\theta \\ \mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right]$

When I rotate a vector $\left[\begin{array}{c}x\\ y\end{array}\right]$ I get

$\left[\begin{array}{c}{x}^{\prime}\\ {y}^{\prime}\end{array}\right]=\left[\begin{array}{c}x\cdot \mathrm{cos}\theta \phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}y\cdot \mathrm{sin}\theta \\ x\cdot \mathrm{sin}\theta \phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}y\cdot \mathrm{cos}\theta \end{array}\right]$

Correct me if I'm wrong, but I thought that column 1 of A $\left[\begin{array}{c}\mathrm{cos}\theta \\ \mathrm{sin}\theta \end{array}\right]$, holds the 'x' values and column 2 holds the 'y' values. What I'm confused about is why does x' contain both an x component and a y component?

and

$T({e}_{2})=T(0,1)=(-\mathrm{sin}\theta ,\mathrm{cos}\theta )$

and

$A=[T({e}_{1})|T({e}_{2})]=\left[\begin{array}{cc}\mathrm{cos}\theta & -\mathrm{sin}\theta \\ \mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right]$

When I rotate a vector $\left[\begin{array}{c}x\\ y\end{array}\right]$ I get

$\left[\begin{array}{c}{x}^{\prime}\\ {y}^{\prime}\end{array}\right]=\left[\begin{array}{c}x\cdot \mathrm{cos}\theta \phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}y\cdot \mathrm{sin}\theta \\ x\cdot \mathrm{sin}\theta \phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}y\cdot \mathrm{cos}\theta \end{array}\right]$

Correct me if I'm wrong, but I thought that column 1 of A $\left[\begin{array}{c}\mathrm{cos}\theta \\ \mathrm{sin}\theta \end{array}\right]$, holds the 'x' values and column 2 holds the 'y' values. What I'm confused about is why does x' contain both an x component and a y component?