Consider the following two bases for R3 : α:={[213],[−101],[31−1]}and β:={[111],[−231],[23−1]} If [x]α=[12−1]αthen find [x]β (that is, express x in the β coordinates).

α={[213],[−101],[31−1]}β={[111],[−231],[23−1]} [x]α=[12−1]α→x=1.[213]+2[−101]−1[311]=[−306] Let x=C1[111]+C2[−231]+C3[23−1]⇒C1−2C2+2C3=−3−(1)C1+3C+2+3C3=0−(2)C1+C2−C3=6−(3) From (3) C3=C1+C2−6 From (1) C1−2C2+2C1+2C2−12=−3⇒3C1=+9⇒C1=3 From (2) C1+3C2+3C1+3C2−18=0⇒4C1+6C2=18⇒6C2=C2 C3=3+1−6=−2 So [−306]=+3[111]+1[−231]−2[23−1] So [x]β=[31−2]

Explained: "Consider the following two bases for R3 :..."

College
Algebra
Linear algebra
Alternate coordinate systems

fortdefruitI

Answered question

2020-11-02

Consider the following two bases for $R^{3}$ :
$α := {[\begin{matrix} 2 \\ 1 \\ 3 \end{matrix}], [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}], [\begin{matrix} 3 \\ 1 \\ - 1 \end{matrix}]} a n d β := {[\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}], [\begin{matrix} - 2 \\ 3 \\ 1 \end{matrix}], [\begin{matrix} 2 \\ 3 \\ - 1 \end{matrix}]}$ If $[x]_{α} = {[\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix}]}_{α} t h e n f i n d [x]_{β}$
(that is, express x in the $β$ coordinates).

Answer & Explanation

Pohanginah

Skilled2020-11-03Added 96 answers

$α = {[\begin{matrix} 2 \\ 1 \\ 3 \end{matrix}], [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}], [\begin{matrix} 3 \\ 1 \\ - 1 \end{matrix}]} β = {[\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}], [\begin{matrix} - 2 \\ 3 \\ 1 \end{matrix}], [\begin{matrix} 2 \\ 3 \\ - 1 \end{matrix}]}$
$[x]_{α} = {[\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix}]}_{α} \to x = 1. [\begin{matrix} 2 \\ 1 \\ 3 \end{matrix}] + 2 [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}] - 1 [\begin{matrix} 3 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} - 3 \\ 0 \\ 6 \end{matrix}]$ Let $x = C_{1} [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] + C_{2} [\begin{matrix} - 2 \\ 3 \\ 1 \end{matrix}] + C_{3} [\begin{matrix} 2 \\ 3 \\ - 1 \end{matrix}] \Rightarrow \begin{matrix} C_{1} - 2 C_{2} + 2 C_{3} = - 3 - (1) \\ C_{1} + 3 C + 2 + 3 C_{3} = 0 - (2) \\ C_{1} + C_{2} - C_{3} = 6 - (3) \end{matrix}$ From (3) $C_{3} = C_{1} + C_{2} - 6$ From (1) $C_{1} - 2 C_{2} + 2 C_{1} + 2 C_{2} - 12 = - 3 \Rightarrow 3 C_{1} = + 9 \Rightarrow C_{1} = 3$ From (2) $C_{1} + 3 C_{2} + 3 C_{1} + 3 C_{2} - 18 = 0 \Rightarrow 4 C_{1} + 6 C_{2} = 18 \Rightarrow 6 C_{2} = C_{2}$
$C_{3} = 3 + 1 - 6 = - 2$ So $[\begin{matrix} - 3 \\ 0 \\ 6 \end{matrix}] = + 3 [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] + 1 [\begin{matrix} - 2 \\ 3 \\ 1 \end{matrix}] - 2 [\begin{matrix} 2 \\ 3 \\ - 1 \end{matrix}]$ So $[x]_{β} = [\begin{matrix} 3 \\ 1 \\ - 2 \end{matrix}]$

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