Let $A$ be the matrix below and define a transformation $T:\mathbb{R}^{3}\to\mathbb{R}^{3}$ by $T(U) = AU$. Let $A$ be the matrix below and define a transformation $T:\mathbb{R}^{3}\to\mathbb{R}^{3}$ by $T(U) = AU$. For each of the vectors $B$ below, find a vector $U$ such that $T$ maps $U$ to $B$, if possible. Otherwise state that there is no such $U$
$A = \pmatrix{5&10&10\\2&7&10\\-1&0&2}$

*

*$b = \pmatrix{0\\3\\2}$

*$b = \pmatrix{-1\\2\\3}$.

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 A: Not going to give you the answer but your general strategy to solving this is to create a "generic" vector $\begin{bmatrix}a\\b\\c\end{bmatrix}$ and see where your transformation $T$ sends it: this will give you some vector $\begin{bmatrix} ac_1\\ bc_2\\ cc_3\end{bmatrix}$. Then set this equal to your target $B$ vector and see what values work! (system of equations).
A: It goes like that:
First we define the variable $\mathrm x=\begin{bmatrix} x\\y\\z\end{bmatrix}$.
Now, we want to solve the equation $A\mathrm x=b$.
That's a linear system of equation, which can be solved using gaussian elimination.
For example, if $b=\begin{bmatrix}0\\3\\2\end{bmatrix}$, the system is
$\begin{matrix}5x+10y+10z=0\\2x+7y+10z=3\\-x+2z=2\end{matrix}$
and after elimination you will get (using wolfram alpha)
$\begin{matrix}x-2z=-2\\y+2z=1\end{matrix}$
Here z is a free variable, and the solution is $\mathrm x=\begin{bmatrix}2t-2\\1-2t\\t\end{bmatrix}$.
If you choose $t=0$, you get $\mathrm x=\begin{bmatrix}-2\\1\\0\end{bmatrix}$.
