Inverse of a linear transformation

**cristi92** · December 30th 2011, 12:00 PM

$A:\mathbb{R}_{2}[x]->\mathbb{R}^{3} , A(P)=(P(-1),P(0),P(1))$
where A is a bijective linear transformation, $\mathbb{R}_{2}[x]$ is the set of polynomials with real coeficients and max degree=2 .
Find $A^{-1}(0,6,0)$

To do this I have to find the matrix of A relative to canonical basis of $\mathbb{R}_{2}[x]$ and $\mathbb{R}^{3}$ , find the inverse of that matrix and then $A^{-1}\begin{pmatrix}0\\ 6\\ 0\end{pmatrix}$ is the answer I'm looking for.

Please tell me if I am right or not.

**ILikeSerena** · December 30th 2011, 12:35 PM

Welcome to MHF, cristi92!

That looks fine to me!

**cristi92** · December 30th 2011, 12:38 PM

Thank you!

**HallsofIvy** · December 30th 2011, 12:55 PM

That would work but since you are NOT asked to find either A or $A^{-1}$ in general, it would seem easier to me to state the problem as "find numbers, a, b, c, so that $P(x)= ax^2+ bx+ c$ , such that $P(-1)= a(-1)^2+ b(-1)+ c= a- b+ c= 0$ , $P(0)= a(0^2)+ b(0)+ c= c= 6$ , $P(1)= a(1^2)+ b(1)+ c= a+ b+ c= 0$ ". In other words solve the three equations a- b+ c= 0, c= 6, a+ b+ c= 0 for a, b, and c.

Of course, if you solve those equations by setting up the matrix of coeffients and row-reducing, you are doing what you describe but those are particularly easy equations to solve.

**cristi92** · December 30th 2011, 12:59 PM

Thank you!

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