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

$B1=\{(1,0)(0,1)\}\to$ standard basis $R^2$ Alternate basis $B2=\{(1,2)(2,-1)\},B2=\{(1,1),(-1,1)\}B2=\{(1,2)(2,1)\},andB2\{(1,2)(-2,1)\})$ a. $(0,5)=a(1,2)+b(-2,1)=(a-2b,2a+b)$
$\Rightarrow a-2b=0$ and 2a + b = 5
$\Rightarrow a=2b\Rightarrow 7b+b=5\Rightarrow 5b=5\Rightarrow b=1thena=2,$
$\therefore (0,5)=1(1,2)+2(-2,1)(1,7)=(1,2)+b(-2,1)=(a-2b,2a+b)\Rightarrow a-2b=1\text{ and }2a+b=7$
$\frac{-2a-ab=-2}{5b=5}\Rightarrow b=1\Rightarrow a=3\therefore (1,7)=3(1,2)+1(-2,1)$ b. $B2=\{(1,2)(2,-1)\}$ is orthogonal since $\left[\begin{array}{c}1\\ 2\end{array}\right]\cdot \left[\begin{array}{c}2\\ -1\end{array}\right]=2-2=0$
$B2=\{(1,1)(1,-1)\}$ is orthogonal since $(\left[\begin{array}{c}1\\ 1\end{array}\right]\cdot \left[\begin{array}{c}1\\ -1\end{array}\right])=1-1=0B2=\{(1,2)(2,1)\}$ is not orthogonal since $\left[\begin{array}{c}1\\ 2\end{array}\right]\cdot \left[\begin{array}{c}2\\ 1\end{array}\right]=2+2=7\ne 0.B2=\{(1,2)(-2,1)\}$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 $Bis{T}_{B2\leftarrow B}=[[b_1{]}_{B2}[b_2{]}_{B2}]$
$b_1=\left[\begin{array}{c}1\\ 0\end{array}\right],b_2=\left[\begin{array}{c}0\\ 1\end{array}\right],B=\{b_1,b_2\}\to$ standard basis $B2=\{(1,2)(2,-1)\}$Then $\left(\begin{array}{c}1\\ 0\end{array}\right)=a\left(\begin{array}{c}1\\ 2\end{array}\right)+b\left(\begin{array}{c}2\\ -1\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}a+2b\\ 2a-b\end{array}\right)\Rightarrow a=\frac{1}{5}b=\frac{2}{5}$
Therefore

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