# Let A and C be the following ordered bases of P3(t): A = (1, 1 + t, 1 + t + t^2, 1 + t + t^2 + t^3) C = (1, -t, t^2 -t^3) Find tha change of coordinate matrix I_{CA}

Let A and C be the following ordered bases of P3(t): $A=\left(1,1+t,1+t+{t}^{2},1+t+{t}^{2}+{t}^{3}\right)$
$C=\left(1,-t,{t}^{2}-{t}^{3}\right)$ Find tha change of coordinate matrix ${I}_{CA}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

irwchh

Solution: $A=\left(1,1+t,1+t+{t}^{2},1+t+{t}^{2}+{t}^{3}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}C=\left(1,-t,{t}^{2},-{t}^{3}\right)$ To find the chance of coordinate matrix ${I}_{CA}$:
$1={a}_{1}\left(1\right)+{a}_{2}\left(1+t\right)+{a}_{3}\left(1+t+{t}^{2}\right)+{a}_{4}\left(1+t+{T}^{2}+{t}^{3}\right)$
$⇒{a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}=1,{a}_{2}+{a}_{3}+{a}_{4}=0,{a}_{3}+{a}_{4}=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{a}_{4}=0$
$⇒{a}_{1}=1$
$-t={b}_{1}\left(1\right)+{b}_{2}\left(1+t\right)+{b}_{3}\left(1+t+{t}_{2}\right)+{b}_{4}\left(1+t+{t}^{2}+{t}^{3}\right)$
$Ri>↔ow{b}_{1}+{b}_{2}+{b}_{3}+{b}_{4}=0,{b}_{2}+{b}_{3}+{b}_{4}=-1,{b}_{3}+{b}_{4}=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{b}_{4}=0$
$⇒{b}_{1}=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{b}_{2}=-1$
${t}^{2}={c}_{1}\left(1\right)+{c}_{2}\left(1+t\right)+{c}_{3}\left(1+t+{t}^{2}\right)+{c}_{4}\left(1+t+{t}^{2}+{t}^{3}\right)$
$⇒{c}_{1}+{c}_{2}+{c}_{3}+{c}_{4}=0,{c}_{2}+{c}_{3}+{c}_{4}=0,{c}_{3}+{c}_{4}=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{c}_{4}=0$
$⇒{c}_{2}=-1,{c}_{3}=1$