Let B and C be the following ordered bases of R3: B=([14−43],[018],[11−2]) C=([11−2],[14−43],[018]) Find the change of coordinate matrix I_{CB}

B=([14−43],[018],[11−2]) C=([11−2],[14−43],[018]) [110101141411−2−438−438−2] [110101011311300−238−2380]−R1+R2→R22R1+R3→R3R2×13→R23−b±b2−4ac2a [11010101131130001010]−23R2+R3→R3R3×970→R9R370 [100001010100001010]−13R3+R2→R2−R2+R1→R1 ICB=[001100010]

Question: "Let B and C be the following ordered..."

College
Algebra
Linear algebra
Alternate coordinate systems

CoormaBak9

Answered question

2020-10-25

Let B and C be the following ordered bases of $R^{3} :$
$B = ([\begin{matrix} 1 \\ 4 \\ - \frac{4}{3} \end{matrix}], [\begin{matrix} 0 \\ 1 \\ 8 \end{matrix}], [\begin{matrix} 1 \\ 1 \\ - 2 \end{matrix}])$
$C = ([\begin{matrix} 1 \\ 1 \\ - 2 \end{matrix}], [\begin{matrix} 1 \\ 4 \\ - \frac{4}{3} \end{matrix}], [\begin{matrix} 0 \\ 1 \\ 8 \end{matrix}])$ Find the change of coordinate matrix I_{CB}

Answer & Explanation

likvau

Skilled2020-10-26Added 75 answers

$B = ([\begin{matrix} 1 \\ 4 \\ - \frac{4}{3} \end{matrix}], [\begin{matrix} 0 \\ 1 \\ 8 \end{matrix}], [\begin{matrix} 1 \\ 1 \\ - 2 \end{matrix}])$
$C = ([\begin{matrix} 1 \\ 1 \\ - 2 \end{matrix}], [\begin{matrix} 1 \\ 4 \\ - \frac{4}{3} \end{matrix}], [\begin{matrix} 0 \\ 1 \\ 8 \end{matrix}])$
$[\begin{matrix} 1 & 1 & 0 & 1 & 0 & 1 \\ 1 & 4 & 1 & 4 & 1 & 1 \\ - 2 & - \frac{4}{3} & 8 & - \frac{4}{3} & 8 & - 2 \end{matrix}]$
$[\begin{matrix} 1 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & \frac{1}{3} & 1 & \frac{1}{3} & 0 \\ 0 & - \frac{2}{3} & 8 & - \frac{2}{3} & 8 & 0 \end{matrix}] \begin{matrix} - R_{1} + R_{2} \to R_{2} \\ 2 R_{1} + R_{3} \to R_{3} \\ R_{2} \times \frac{1}{3} \to \frac{R_{2}}{3} \end{matrix} \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$[\begin{matrix} 1 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & \frac{1}{3} & 1 & \frac{1}{3} & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{matrix}] \begin{matrix} - \frac{2}{3} R_{2} + R_{3} \to R_{3} \\ R_{3} \times \frac{9}{70} \to \frac{R_{9 R_{3}}}{70} \end{matrix}$
$[\begin{matrix} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{matrix}] \begin{matrix} - \frac{1}{3} R_{3} + R_{2} \to R_{2} \\ - R_{2} + R_{1} \to R_{1} \end{matrix}$
$I_{C B} = [\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}]$

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