gorgeousgen9487

Answered

2022-07-06

$A=\left(\begin{array}{cc}k& -2\\ 1-k& k\end{array}\right)\text{, where k is a constant}$

$\text{A transformation}T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}\text{is represented by the matrix A.}$

$\text{Find the value of k for which the line}y=2x\text{is mapped onto itself under T.}$

$\text{A transformation}T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}\text{is represented by the matrix A.}$

$\text{Find the value of k for which the line}y=2x\text{is mapped onto itself under T.}$

Answer & Explanation

Valeria Wolfe

Expert

2022-07-07Added 11 answers

The line $y=2x$ is mapped to itself if we have that for each point $(x,y)$ on the line $T(x,y)$ also lies on the line. We needn't have $T(x,2x)=(x,2x)$ for all $x$, it suffices to have $T(x,2x)={\textstyle (}\varphi (x),2\varphi (x){\textstyle )}$ for all $x$ and some function $\varphi $.

Having this in mind, we look at $T(x,2x)={\textstyle (}x(k-4),x(k+1){\textstyle )}$. We must have $2(k-4)=k+1$ for $(x(k-1),x(k+1))$ to lie on our line. This gives $k=9$

Having this in mind, we look at $T(x,2x)={\textstyle (}x(k-4),x(k+1){\textstyle )}$. We must have $2(k-4)=k+1$ for $(x(k-1),x(k+1))$ to lie on our line. This gives $k=9$

racodelitusmn

Expert

2022-07-08Added 5 answers

On the same line (!) of thought: the line $\phantom{\rule{thinmathspace}{0ex}}l:y=2x\phantom{\rule{thinmathspace}{0ex}}$ is the same as the vector space $\phantom{\rule{thinmathspace}{0ex}}\mathrm{Span}\{(1,2)\}\le {\mathbb{R}}^{2}$ , or if you prefer: $\phantom{\rule{thinmathspace}{0ex}}l:\{(r,2r)\phantom{\rule{thinmathspace}{0ex}}/\phantom{\rule{thinmathspace}{0ex}}r\in \mathbb{R}\}\phantom{\rule{thinmathspace}{0ex}}$, and then what we really want to happen is

$\left(\begin{array}{cc}k& -2\\ 1-k& k\end{array}\right)\left(\begin{array}{c}1\\ 2\end{array}\right)=\left(\begin{array}{c}r\\ 2r\end{array}\right)\u27f9\begin{array}{}k=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}r+4\\ k=2r-1\end{array}$

so $\phantom{\rule{thinmathspace}{0ex}}r+4=2r-1\u27f9r=5\phantom{\rule{thinmathspace}{0ex}}$ and thus $\phantom{\rule{thinmathspace}{0ex}}k=9\phantom{\rule{thinmathspace}{0ex}}$

$\left(\begin{array}{cc}k& -2\\ 1-k& k\end{array}\right)\left(\begin{array}{c}1\\ 2\end{array}\right)=\left(\begin{array}{c}r\\ 2r\end{array}\right)\u27f9\begin{array}{}k=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}r+4\\ k=2r-1\end{array}$

so $\phantom{\rule{thinmathspace}{0ex}}r+4=2r-1\u27f9r=5\phantom{\rule{thinmathspace}{0ex}}$ and thus $\phantom{\rule{thinmathspace}{0ex}}k=9\phantom{\rule{thinmathspace}{0ex}}$

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