Answer to Question #293865 in Linear Algebra for Himanshu

Question

Express v = t2+4t-3 in P(t) as a linear combination of the

polynomials p =t2−2t+5,p =2t2 −3tand p =t+1.

Answer 1

$v=t^2+4t-3\\ p_1=t^2-2t+5\\ p_2=2t^2-3t\\ p_3=t+1\\ v=Ap_1+Bp_2+Cp_3$

$t^2+4t-3=\\ =A(t^2-2t+5)+B(2t^2-3t)+C(t+1)\\$

$t^2: A+2B=1\\ t:-2A-3B+C=4\\ t^0: 5A+C=-3$

$\begin{pmatrix} 1 & 2&0&|&1 \\ -2&-3&1&|&4\\ 5&0&1&|&-3 \end{pmatrix}\mapsto$

2r+1r(2)

3r+1r(-5)

$\begin{pmatrix} 1 & 2&0&|&1 \\ 0&1&1&|&6\\ 0&-10&1&|&-8 \end{pmatrix}\mapsto$

3r+2r(10)

$\begin{pmatrix} 1 & 2&0&|&1 \\ 0&1&1&|&6\\ 0&0&11&|&52 \end{pmatrix}\mapsto$

3r($\frac{1}{11}$ )

$\begin{pmatrix} 1 & 2&0&|&1 \\ 0&1&1&|&6\\ 0&0&1&|&\frac{52}{11} \end{pmatrix}\mapsto$

2r+3r(-1)

$\begin{pmatrix} 1 & 2&0&|&1 \\ 0&1&0&|&\frac{14}{11}\\ 0&0&1&|&\frac{52}{11} \end{pmatrix}\mapsto$

1r+2r(-2)

$\begin{pmatrix} 1 & 0&0&|&-\frac{17}{11} \\ 0&1&0&|&\frac{14}{11}\\ 0&0&1&|&\frac{52}{11} \end{pmatrix}$

$A=-\frac{17}{11}, B=\frac{14}{11}, C=\frac{52}{11}\\ v=t^2+4t-3=-\frac{17}{11}(t^2-2t+5)+\\ +\frac{14}{11}(2t^2-3t)+\frac{52}{11}(t+1)$