e1s2kat26

2020-11-01

Let $\gamma =\left\{{t}^{2}-t+1,t+1,{t}^{2}+1\right\}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\beta =\left\{{t}^{2}+t+4,4{t}^{2}-3t+2,2{t}^{2}+3\right\}be\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}deredbasesf\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{P}_{2}\left(R\right).$ Find the change of coordinate matrix Q that changes

i1ziZ

Let ${t}^{2}+t+4={\gamma }_{1}\left({t}^{2}-t+1\right)+{\gamma }_{2}\left(t=1\right)+{\gamma }_{3}\left({t}^{2}+1\right)$
$⇒{\gamma }_{1}+{\gamma }_{3}=1,-{\gamma }_{1}+{\gamma }_{2}=1,{\gamma }_{1}+{\gamma }_{2}+{\gamma }_{3}=4$
$⇒1+{\gamma }_{2}=4$
$⇒{\gamma }_{2}=3$
$-{\gamma }_{1}+{\gamma }_{2}=1⇒-{\gamma }_{1}+3=1⇒{\gamma }_{1}=2$
${\gamma }_{1}+{\gamma }_{3}=1⇒2+{\gamma }_{3}=1⇒{\gamma }_{3}=-1$ Let $4{t}^{2}-3r+2={\gamma }_{1}\left({t}^{2}-t+1\right)+{\gamma }_{2}\left(t+1\right)+{\gamma }_{3}\left({t}^{2}+1\right)$
$⇒{\gamma }_{1}+{\gamma }_{3}=4,-{\gamma }_{1}+{\gamma }_{2}=-3,{\gamma }_{1}+{\gamma }_{2}+{\gamma }_{3}=2$
$⇒4+{\gamma }_{2}=2$
$⇒{\gamma }_{2}=-2$
$-{\gamma }_{1}+{\gamma }_{2}=-3⇒-{\gamma }_{1}-2=-3⇒{\gamma }_{1}=1$
${\gamma }_{1}+{\gamma }_{3}=4⇒1+{\gamma }_{3}=4⇒{\gamma }_{3}=3$ Let $2{t}^{2}+3={\gamma }_{1}\left({t}^{2}-t+1\right)+{\gamma }_{2}\left(t+1\right)+{\gamma }_{3}\left({t}^{2}+1\right)$
$⇒{\gamma }_{1}+{\gamma }_{3}=2,-{\gamma }_{1}+{\gamma }_{2}=0,{\gamma }_{1}+{\gamma }_{2}+{\gamma }_{3}=3$
$⇒2+{\gamma }_{2}=3$
$⇒{\gamma }_{2}=1$
$-{\gamma }_{1}+{\gamma }_{2}=0⇒-{\gamma }_{1}+1=0Ri>↔ow{\gamma }_{1}=1$
${\gamma }_{1}+{\gamma }_{3}=2⇒1+{\gamma }_{3}=2⇒{\gamma }_{3}=1$

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