$\begingroup$- What is $T(x,y,z)$
What I tried: $$T(x,y,z)= \begin{pmatrix} 1&1&1|0&1&2\\ 1&0&1|1&1&1\\ 0&0&1|1&2&3\\ \end{pmatrix} $$ Which reduces to \begin{pmatrix} 1&0&0|&0&-1&-2\\ 0&1&0|&-1&0&1\\ 0&0&1|&1&2&3\\ \end{pmatrix}
So that would mean $T(x,y,z)=(-y-2z,-x+z,x+2y+3z)$
Or did I do it completely wrong?
$\endgroup$4 Answers
$\begingroup$$$T(1,0,0)=T\left((1,0,1)-(0,0,1)\right)=T(1,0,1)-T(0,0,1)=(1,1,1)-(1,2,3)=(0,-1,-2)$$ $$T(0,1,0)=T(1,1,1)-T(1,0,0)-T(0,0,1)=(0,1,2)-(0,-1,-2)-(1,2,3)=(-1,0,1)$$ $$T(0,0,1)=(1,2,3)$$ Then $$T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1)$$
Can you take it from here?
$\endgroup$$\begingroup$Let $v_1$, $v_2$, and $v_3$ be the respective columns of $$ P= \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} $$ and note that $$ P^{-1}= \begin{bmatrix} \color{red}{0}&\color{blue}{1}&\color{green}{0}\\ \color{red}{1}&\color{blue}{-1}&\color{green}{0}\\ \color{red}{-1}&\color{blue}{0}&\color{green}{1} \end{bmatrix} $$ Since $PP^{-1}=I$, it follows that \begin{array}{rcrcrcrcrcr} \color{red}{0}\,v_1&+&\color{red}{1}\,v_2&+&\color{red}{-1}\,v_3 &=& e_1\\ \color{blue}{1}\,v_1&+&\color{blue}{-1}\,v_2&+&\color{blue}{0}\,v_3&=&e_2\\ \color{green}0\,v_1&+&\color{green}0\,v_2&+&\color{green}1\,v_3&=&e_3 \end{array} where $\{e_1,e_2,e_3\}$ is the standard basis for $\Bbb R^3$. Applying $T$ to these equations gives \begin{array}{rcrcrcrcrcr} T(e_1) &=& \color{red}{0}\,T(v_1)&+&\color{red}{1}\,T(v_2)&+&\color{red}{-1}\,T(v_3)\\ &=& \color{red}{0}\,\langle0,1,2\rangle &+& \color{red}{1}\,\langle1,1,1\rangle &+& \color{red}{-1}\,\langle0,0,1\rangle \\ & = & \langle1,1,0\rangle\\ T(e_2) &=&\color{blue}{1}\,T(v_1)&+&\color{blue}{-1}\,T(v_2)&+&\color{blue}{0}\,T(v_3)\\ &=& \color{blue}{1}\,\langle0,1,2\rangle &+& \color{blue}{-1}\,\langle1,1,1\rangle &+& \color{blue}{0}\,\langle0,0,1\rangle \\ & = & \langle -1,0,1\rangle\\ T(e_3)&=&\color{green}0\,T(v_1)&+&\color{green}0\,T(v_2)&+&\color{green}1\,T(v_3) \\ &=& \color{green}{0}\,\langle0,1,2\rangle &+& \color{green}{0}\,\langle1,1,1\rangle &+& \color{green}{1}\,\langle0,0,1\rangle \\ & = & \langle0,0,1\rangle \end{array} Hence \begin{align*} T(x,y,z) &= T(xe_1+ye_2+ze_3) \\ &= xT(e_1)+yT(e_2)+zT(e_3) \\ &= x\langle1,1,0\rangle+y\langle-1,0,1\rangle+z\langle0,0,1\rangle \\ &= \langle x-y,x,y+z\rangle \end{align*}
$\endgroup$$\begingroup$Let $T(x,y,z)=A\begin{bmatrix}x\\y\\z\end{bmatrix}$.
Then $A\begin{bmatrix}1&1&0\\1&0&0\\1&1&1\end{bmatrix}=\begin{bmatrix}0&1&1\\1&1&2\\2&1&3\end{bmatrix}$, so
$A=\begin{bmatrix}0&1&1\\1&1&2\\2&1&3\end{bmatrix}\begin{bmatrix}1&1&0\\1&0&0\\1&1&1\end{bmatrix}^{-1}=\begin{bmatrix}0&1&1\\1&1&2\\2&1&3\end{bmatrix}\begin{bmatrix}0&1&0\\1&-1&0\\-1&0&1\end{bmatrix}=\begin{bmatrix}0&-1&1\\-1&0&2\\-2&1&3\end{bmatrix}$
and $T(x,y,z)=A\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-y+z\\ -x+2z\\ -2x+y+3z\end{bmatrix}$.
$\endgroup$$\begingroup$$$\left[\begin{array}{ccc|ccc} 1&0&0&0&-1&-2\\ 0&1&0&-1&0&1\\ 0&0&1&1&2&3\\ \end{array}\right]$$
You are correct up to this point, but you need to know what the above blurb represents. If you set up the system:
$$[1 ~~ 1 ~~ 1]M=[0 ~~ 1 ~~ 2]$$ $$[1 ~~ 0 ~~ 1]M=[1 ~~ 1 ~~ 1]$$ $$[0 ~~ 0 ~~ 1]M=[1 ~~ 2 ~~ 3]$$
Then you get the above system when you row reduce. So your result should be:
$$T(x,y,z) = [x ~~ y ~~ z]\begin{bmatrix}0 & -1 & -2 \\ -1 & 0 & 1 \\ 1 & 2 & 3 \end{bmatrix}$$
like you set up the original equations, it is not
$$T(x,y,z) = \begin{bmatrix}0 & -1 & -2 \\ -1 & 0 & 1 \\ 1 & 2 & 3 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} \tag{Wrong}$$
which is where your solution got off course. When you set up the system, you treated the inputs and outputs as row vectors, so you must do that in the final result as well.
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