Answer to Question #222357 in Linear Algebra for Muna_N

Question

Express V= 3t² + 7t + -4 as a linear combination of the polynomials

P1= t² + 2t + 3

P2= 2t² + 3t + 7

P3 = 3t² + 5t + 6

Answer 1

Let us express $V= 3t^2+ 7t -4$ as a linear combination of the polynomials

$P_1= t^2 + 2t + 3,\ P_2= 2t^2 + 3t + 7,\ P_3 = 3t^2 + 5t + 6.$

Let $V=aP_1+bP_2+cP_3.$

Then

$3t^2+ 7t -4= a(t^2 + 2t + 3)+b( 2t^2 + 3t + 7)+c( 3t^2 + 5t + 6).$

It follows that

$3t^2+ 7t -4= at^2 + 2at + 3a+2bt^2 + 3bt + 7b+3ct^2 + 5ct + 6c,$

and hence

$3t^2+ 7t -4= (a+2b+3c)t^2 + (2a+3b+5c)t + (3a+ 7b + 6c).$

Therefore, we have the following system:

$\begin{cases} a+2b+3c=3\\ 2a+3b+5c=7\\ 3a+ 7b + 6c=-4 \end{cases}$

Let us multiply the first equation by $-2$ and add to the second equation, also multiply the first equation by $-3$ and add to the third equation. Thgen we have the system:

$\begin{cases} a+2b+3c=3\\ -b-c=1\\ b -3c=-13 \end{cases}$

Let us add the last to equation. Then we get the system:

$\begin{cases} a=3-2b-3c\\ -b-c=1\\ -4c=-12 \end{cases}$

The last system is equivalent to the system:

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

We conclude that $V=2P_1-4P_2+3P_3.$