 # Given the elow bases for R^2 and the point at the specified coordinate in the standard basis as below, (40 points) Wribreeminsl 2020-10-18 Answered

Given the elow bases for ${R}^{2}$ and the point at the specified coordinate in the standard basis as below, (40 points)

$\left(B1=\left\{\left(1,0\right),\left(0,1\right)\right\}$&

$B2=\left(1,2\right),\left(2,-1\right)\right\}$(1, 7) = ${3}^{\ast }\left(1,2\right)-\left(2,1\right)$
$B2=\left(1,1\right),\left(-1,1\right)\left(3,7={5}^{\ast }\left(1,1\right)+{2}^{\ast }\left(-1,1\right)$

$\left(8,10\right)={4}^{\ast }\left(1,2\right)+{2}^{\ast }\left(2,1\right)$
B2 = (1, 2), (-2, 1) (0, 5) =
(1, 7) =

a. Use graph technique to find the coordinate in the second basis. (10 points) b. Show that each basis is orthogonal. (5 points) c. Determine if each basis is normal. (5 points) d. Find the transition matrix from the standard basis to the alternate basis. (15 points)

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it lamusesamuset

$B1=\left\{\left(1,0\right)\left(0,1\right)\right\}\to$ standard basis ${R}^{2}$ Alternate basis $B2=\left\{\left(1,2\right)\left(2,-1\right)\right\},B2=\left\{\left(1,1\right),\left(-1,1\right)\right\}B2=\left\{\left(1,2\right)\left(2,1\right)\right\},andB2\left\{\left(1,2\right)\left(-2,1\right)\right\}\right)$ a. $\left(0,5\right)=a\left(1,2\right)+b\left(-2,1\right)=\left(a-2b,2a+b\right)$
$⇒a-2b=0$ and 2a + b = 5
$⇒a=2b⇒7b+b=5⇒5b=5⇒b=1thena=2,$

$\frac{-2a-ab=-2}{5b=5}⇒b=1⇒a=3\therefore \left(1,7\right)=3\left(1,2\right)+1\left(-2,1\right)$
b. $B2=\left\{\left(1,2\right)\left(2,-1\right)\right\}$ is orthogonal since $\left[\begin{array}{c}1\\ 2\end{array}\right]\cdot \left[\begin{array}{c}2\\ -1\end{array}\right]=2-2=0$
$B2=\left\{\left(1,1\right)\left(1,-1\right)\right\}$
is orthogonal since $\left(\left[\begin{array}{c}1\\ 1\end{array}\right]\cdot \left[\begin{array}{c}1\\ -1\end{array}\right]\right)=1-1=0B2=\left\{\left(1,2\right)\left(2,1\right)\right\}$ is not orthogonal since $\left[\begin{array}{c}1\\ 2\end{array}\right]\cdot \left[\begin{array}{c}2\\ 1\end{array}\right]=2+2=7\ne 0.B2=\left\{\left(1,2\right)\left(-2,1\right)\right\}$is orthogonal c. The vector $U\in {R}^{2}$ is normal of $null=1\cdot B$ at every vector in B2 is not normal. They are unit vectors X $\therefore$ All bases B2 are not normal X. d. The transition matrix from B to $Bis{T}_{B2←B}=\left[\left[{b}_{1}{\right]}_{B2}\left[{b}_{2}{\right]}_{B2}\right]$
${b}_{1}=\left[\begin{array}{c}1\\ 0\end{array}\right],{b}_{2}=\left[\begin{array}{c}0\\ 1\end{array}\right],B=\left\{{b}_{1},{b}_{2}\right\}\to$
standard basis $B2=\left\{\left(1,2\right)\left(2,-1\right)\right\}$Then $\left(\begin{array}{c}1\\ 0\end{array}\right)=a\left(\begin{array}{c}1\\ 2\end{array}\right)+b\left(\begin{array}{c}2\\ -1\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}a+2b\\ 2a-b\end{array}\right)⇒a=\frac{1}{5}b=\frac{2}{5}$
Therefore