# Product of linear mapping

Matrix $A=\Bigg| \begin{matrix} 2 & 0 & 4 \\ -1 & -1 & -2 \\ \end{matrix}\Bigg|$ is a matrix of linear mapping $l: R^3\to R^2$ with the respect to bases $B = \{(1,0,0),(0,-1,1),(1,1,1)\}$ and $B' = \{(1,1),(1,2)\}$.

Find $l(1,2,3)$.

My result is $(7,-4)$, but I'm afraid I did some mistake. Could you, please, write your solution?

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Why don't you just check it? How would you go about checking it? – k.stm Jan 20 '13 at 10:57
How am I supposed to check it? – user50222 Jan 20 '13 at 11:04

(1) Write $\,(1,2,3)\,$ as a linear combination of the given basis of $\,\Bbb R^3\,$:

$$(1,2,3)=a(1,0,0)+b(0,-1,1)+c(1,1,1)\,\,\,,\,\,a,b,c,\in\Bbb R$$

For example, $\,c=5/2\,$ ...

(2) Now just calculate

$$A\begin{pmatrix}a\\b\\c\end{pmatrix}$$

and get your vector...which is going to be given expressed in the given basis of $\,\Bbb R^2\,$, of course.

Addendum, @user50222: So you can use the coefficients of this vector to linearly combine the base vectors from $B'$ to a vector which should be … $(7,-4)$. – k.stm Jan 20 '13 at 12:08