Question

Express V= 2t² + 5t + -5 as a linear combination of the polynomials
P1= t² + 2t + 1
P2= 2t + 5t + 4
P3 = t² + 3t + 6

Answer 1

We assume that $P_2=2t^2+5t+4$. $V$ can be expressed in the following way: $V=c_1P_1+c_2P_2+c_3P_3$ , $c_1,c_2,c_3\in \mathbb{C}$ . Coefficients satisfy the following system:

$$\left\{\begin{matrix} c_1+2c_2+c_3=2 \\ 2c_1+5c_2+3c_3=5 \\ c_1+4c_2+6c_3=-5 \end{matrix}\right.$$

We multiply the first equation by $(-2)$ and add to the second one. Then, we multiply the first equation by $(-1)$ and add to the third one. As a result, we obtain we eliminate $c_1$ from all equations and obtain the system for $c_2,c_3:$

$\left\{\begin{matrix} c_2+c_3=1 \\ 2c_2+5c_3=-7 \end{matrix}\right.$

We multiply the first equation by $(-2)$ and add to the second one. We obtain that $c_3=-3$ . Then, from the first equation of the latter system we get: $c_2=4$. From the first system we get $c_1=-3$ . Thus, the solution is: $c_1=-3,c_2=4,c_3=-3$ . The answer is: $V=-3P_1+4P_2-3P_3$ .