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europaparksn

Answered question

2021-11-18

Find the coordinate vector of p relative to the basis

S = {P 1, P 2, p 3} for P 2 . (b) p = 2 - x + x^{2}; p_{1} = 1 + x, p_{2} = 1 + x^{2}, p_{3} = x + x^{2}

Answer & Explanation

Louis Smith

Beginner2021-11-19Added 14 answers

Step 1
We have
$p = 2 - x + x^{2}, p_{1} = 1 + x, p_{2} = 1 + x^{2}, p_{3} = x + x^{2}$
According to Definition 2,fing the the coordinate vector of v relative to the basis $S = {v_{1}, v_{2}, v_{3}}$
$a (1 + x) + b (1 + x^{2}) + c (x + x^{2}) = 2 - x + x^{2}$
$a + a x + b + b x^{2} + c x + c x^{2} = 2 - x + x^{2}$
$a + b + x (a + c) + x^{2} (b + c) = 2 - x + x^{2}$
The two polynomials are the same if they are equal in the coefficients of the corresponding degrees, then the corresponding system of equations is
$a + b = 2$ (1)
$a + c = - 1$ (2)
$b + c = 1$ (3)
To find a,b,c use Gauss-Jordan method
$[\begin{matrix} 1 & 1 & 0 & | & 2 \\ 1 & 0 & 1 & | & - 1 \\ 0 & 1 & 1 & | & 1 \end{matrix}] \begin{matrix} R_{2} - R_{1} \to R_{2} \\ R_{3} + R_{2} \to R_{3} \end{matrix} \sim [\begin{matrix} 1 & 1 & 0 & | & 2 \\ 0 & - 1 & 1 & | & - 3 \\ 0 & 0 & 2 & | & - 2 \end{matrix}]$
$[\begin{matrix} 1 & 1 & 0 & | & 2 \\ 0 & - 1 & 1 & | & - 3 \\ 0 & 0 & 2 & | & - 2 \end{matrix}] \begin{matrix} \frac{1}{3} R_{3} \to R_{3} \\ R_{2} - R_{3} \to R_{2} \end{matrix} \sim [\begin{matrix} 1 & 1 & 0 & | & 2 \\ 0 & - 1 & 0 & | & - 2 \\ 0 & 0 & 1 & | & - 1 \end{matrix}]$
$[\begin{matrix} 1 & 1 & 0 & | & 2 \\ 0 & - 1 & 0 & | & - 2 \\ 0 & 0 & 1 & | & - 1 \end{matrix}] \begin{matrix} - R_{2} \to R_{2} \\ R_{1} - R_{2} \to R_{1} \end{matrix} \sim [\begin{matrix} 1 & 0 & 0 & | & 0 \\ 0 & 1 & 0 & | & - 2 \\ 0 & 0 & 1 & | & - 1 \end{matrix}]$
$H e n c e, (p) s = (0, 2, - 1)$
Result $(p) s = (0, 2, - 1)$

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