# Given the full and correct answer the two bases of B1

Given the full and correct answer the two bases of $B1=\left\{\left(\begin{array}{c}2\\ 1\end{array}\right),\left(\begin{array}{c}3\\ 2\end{array}\right)\right\}$
${B}_{2}=\left\{\left(\begin{array}{c}3\\ 1\end{array}\right),\left(\begin{array}{c}7\\ 2\end{array}\right)\right\}$
find the change of basis matrix from ${B}_{1}\to {B}_{2}$ and next use this matrix to covert the coordinate vector
${\stackrel{\to }{v}}_{{B}_{1}}=\left(\begin{array}{c}2\\ -1\end{array}\right)$ of v to its coodirnate vector
${\stackrel{\to }{v}}_{{B}_{2}}$

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${B}_{1}=\left\{\left[\begin{array}{c}2\\ 1\end{array}\right],\left[\begin{array}{c}3\\ 2\end{array}\right]\right\}and{B}_{2}=\left\{\left[\begin{array}{c}3\\ 1\end{array}\right],\left[\begin{array}{c}7\\ 2\end{array}\right]\right\}$ Consider $\left[{B}_{2}:{B}_{1}\right]=\left[\begin{array}{cccc}3& 7& 2& 3\\ 1& 2& 1& 2\end{array}\right]$
$\sim \left[\begin{array}{cccc}1& 2& 1& 2\\ 3& 7& 2& 3\end{array}\right]\left({R}_{1}↔{R}_{2}\right)$
$\sim \left[\begin{array}{cccc}1& 2& 1& 2\\ 0& 1& -1& -3\end{array}\right]\left({R}_{1}\to {R}_{2}-3{R}_{1}\right)$
$\sim \left[\begin{array}{cccc}1& 0& 3& 8\\ 0& 1& -1& -3\end{array}\right]\left({R}_{1}↔{R}_{1}-2{R}_{2}\right)$
$\therefore$ change of basis matrix from
${B}_{1}\to {B}_{2}is$
$P=\left[\begin{array}{cc}3& 8\\ -1& -3\end{array}\right]$
$men\left[v{\right]}_{{B}_{1}}=\left[\begin{array}{c}2\\ -1\end{array}\right]$
$men\left[v{\right]}_{{B}_{2}}=P\left[v{\right]}_{{B}_{1}}=\left[\begin{array}{cc}3& 8\\ -1& -3\end{array}\right]\left[\begin{array}{c}2\\ -1\end{array}\right]$
$\therefore \left[v{\right]}_{{B}_{2}}=\left[\begin{array}{c}-2\\ 1\end{array}\right]$