Answer to Question #211641 in Linear Algebra for Muhammad Zubair

Question

Express the following polynomial as a linear combination of the polynomials.

P=t²+ 4t-3

P₁=t²+2t+5

P₂=2t^2-3t

P₃=t+3

Answer 1

Let us express the following polynomial as a linear combination of the polynomials.

$P=aP_1+bP_2+cP_3$

It follows that

$t^2+ 4t-3= a(t^2+2t+5)+b(2t^2-3t)+c(t+3)$,

and hence

$t^2+ 4t-3= (a+2b)t^2+(2a-3b+c)t+(5a+3c)$

We get the following system:

$\begin{cases} a+2b=1\\2a-3b+c=4\\5a+3c=-3\end{cases}$

$\begin{cases} a=1-2b\\2(1-2b)-3b+c=4\\5(1-2b)+3c=-3\end{cases}$

$\begin{cases} a=1-2b\\-7b+c=2\\-10b+3c=-8\end{cases}$

$\begin{cases} a=1-2b\\c=2+7b\\-10b+3(2+7b)=-8\end{cases}$

$\begin{cases} a=1-2b\\c=2+7b\\11b=-14\end{cases}$

$\begin{cases} a=\frac{39}{11}\\c=-\frac{76}{11}\\b=-\frac{14}{11}\end{cases}$

We conclude that

$P=\frac{39}{11}P_1-\frac{14}{11}P_2-\frac{76}{11}P_3$