Given the full and correct answer the two bases of B1

Question

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

Answer 1

$B_{1} = {[\begin{matrix} 2 \\ 1 \end{matrix}], [\begin{matrix} 3 \\ 2 \end{matrix}]} a n d B_{2} = {[\begin{matrix} 3 \\ 1 \end{matrix}], [\begin{matrix} 7 \\ 2 \end{matrix}]}$ Consider $[B_{2} : B_{1}] = [\begin{matrix} 3 & 7 & 2 & 3 \\ 1 & 2 & 1 & 2 \end{matrix}]$
$\sim [\begin{matrix} 1 & 2 & 1 & 2 \\ 3 & 7 & 2 & 3 \end{matrix}] (R_{1} \leftrightarrow R_{2})$
$\sim [\begin{matrix} 1 & 2 & 1 & 2 \\ 0 & 1 & - 1 & - 3 \end{matrix}] (R_{1} \to R_{2} - 3 R_{1})$
$\sim [\begin{matrix} 1 & 0 & 3 & 8 \\ 0 & 1 & - 1 & - 3 \end{matrix}] (R_{1} \leftrightarrow R_{1} - 2 R_{2})$
$∴$ change of basis matrix from
$B_{1} \to B_{2} i s$
$P = [\begin{matrix} 3 & 8 \\ - 1 & - 3 \end{matrix}]$
$m e n [v]_{B_{1}} = [\begin{matrix} 2 \\ - 1 \end{matrix}]$
$m e n [v]_{B_{2}} = P [v]_{B_{1}} = [\begin{matrix} 3 & 8 \\ - 1 & - 3 \end{matrix}] [\begin{matrix} 2 \\ - 1 \end{matrix}]$
$∴ [v]_{B_{2}} = [\begin{matrix} - 2 \\ 1 \end{matrix}]$

Given the full and correct answer the two bases of B1

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