 # Consider the following two bases for R^3 fortdefruitI 2020-11-02 Answered

Consider the following two bases for ${R}^{3}$ :
If
(that is, express x in the $\beta$ coordinates).

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$\alpha =\left\{\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right],\left[\begin{array}{c}3\\ 1\\ -1\end{array}\right]\right\}\beta =\left\{\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],\left[\begin{array}{c}-2\\ 3\\ 1\end{array}\right],\left[\begin{array}{c}2\\ 3\\ -1\end{array}\right]\right\}$
$\left[x{\right]}_{\alpha }={\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]}_{\alpha }\to x=1.\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right]+2\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]-1\left[\begin{array}{c}3\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}-3\\ 0\\ 6\end{array}\right]$ Let $x={C}_{1}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]+{C}_{2}\left[\begin{array}{c}-2\\ 3\\ 1\end{array}\right]+{C}_{3}\left[\begin{array}{c}2\\ 3\\ -1\end{array}\right]⇒\begin{array}{c}{C}_{1}-2{C}_{2}+2{C}_{3}=-3-\left(1\right)\\ {C}_{1}+3C+2+3{C}_{3}=0-\left(2\right)\\ {C}_{1}+{C}_{2}-{C}_{3}=6-\left(3\right)\end{array}$ From (3) ${C}_{3}={C}_{1}+{C}_{2}-6$ From (1) ${C}_{1}-2{C}_{2}+2{C}_{1}+2{C}_{2}-12=-3⇒3{C}_{1}=+9⇒{C}_{1}=3$ From (2) ${C}_{1}+3{C}_{2}+3{C}_{1}+3{C}_{2}-18=0⇒4{C}_{1}+6{C}_{2}=18⇒6{C}_{2}={C}_{2}$
${C}_{3}=3+1-6=-2$ So $\left[\begin{array}{c}-3\\ 0\\ 6\end{array}\right]=+3\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]+1\left[\begin{array}{c}-2\\ 3\\ 1\end{array}\right]-2\left[\begin{array}{c}2\\ 3\\ -1\end{array}\right]$ So $\left[x{\right]}_{\beta }=\left[\begin{array}{c}3\\ 1\\ -2\end{array}\right]$