Find $a\in\mathbb R$, if it exists, so that $T(2,0,1)=(1,-\frac 52)$ Let $$T:\mathbb R^3\to \mathbb R^2\mid M_{B_1B_2}=\begin{pmatrix}a&1&2\\-1&0&\color{red}1\end{pmatrix}$$ a linear transformation and the basis $$B_1=\{(1,0,0),(0,-3,1),(0,0,-2)\},\qquad B_2=\{(2,0),(0,-1)\}.$$
Find $a\in\mathbb R$, if it exists, so that $T(2,0,1)=\left(1,-\dfrac 52\right).$

I did a similar exercise and the steps I did are the same:
$$\left(\begin{array}{cc:c}2&0&1\\0&-1&-5/2\end{array}\right)\sim \left(\begin{array}{cc:c}1&0&1/2\\0&1&5/2\end{array}\right).$$ Then
$$\left(\begin{array}{ccc:c}1&0&0&2\\0&-3&0&0\\0&1&-2&1\end{array}\right)\sim \left(\begin{array}{ccc:c}1&0&0&2\\0&1&0&0\\0&0&1&-1/2\end{array}\right)$$
so we need to do
$$\begin{array}{cccccc}
M_{B_1B_2}&\cdot &[v]_{B_1}&=&[T(v)]_{B_2}& \\
\begin{pmatrix}a&1&2\\-1&0&\color{red}1\end{pmatrix}&\cdot&\begin{pmatrix}2\\0\\-1/2\end{pmatrix}&=&\begin{pmatrix}\color{red}{1/2}\\5/2\end{pmatrix}&\\
&\begin{pmatrix}2a-1\\-5/2\end{pmatrix}&&=&\begin{pmatrix}\color{red}{1/2}\\5/2\end{pmatrix}&\Rightarrow \begin{cases}2a-1&=&\color{red}{1/2}\\-5/2&=&5/2\end{cases}
\end{array}$$
but $$\boxed{-\dfrac 52\neq \dfrac 52\quad\wedge\quad a=\dfrac 34}$$ and the answer says $\boxed{a=1}\;\text{(apparently without contradictions)}$.
So I came to an absurd and the value of $a$ gives me different, what can be said of the exercise? Is it wrong or am I wrong?
Thanks!
$\color{red}{\text{Edited in red.}}$
 A: Note that $$[(2,0,1)]_{B_1}=(2,0,\frac{-1}{2})_{B_1}$$ because $$(2,0,1)=2(1,0,0)+0(0,-3,1)+\frac{-1}{2}(0,0,-2)$$, thus, $$T(2,0,1)=(2a-1,\frac{-5}{2})$$ ,take a=1
A: Find the matrix of $T$ with respect to the canonical bases of $\mathbb{R}^{3}$ and $\mathbb{R}^2$, which is
$$
\begin{bmatrix}
2 & 0 \\
0 & -1
\end{bmatrix}
\begin{bmatrix}
a & 1 & 2 \\
-1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & -3 & 0 \\
0 & 1 & -2
\end{bmatrix}^{-1}
$$
In order to find the required inverse, one can use Gaussian elimination:
$$
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & -3 & 0 & 0 & 1 & 0 \\
0 & 1 & -2 & 0 & 0 & 1
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & -2 & 0 & 0 & 1 \\
0 & -3 & 0 & 0 & 1 & 0
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & -2 & 0 & 0 & 1 \\
0 & 0 & -6 & 0 & 1 & 3
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & -2 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & -1/6 & -1/2
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & -1/3 & 0 \\
0 & 0 & 1 & 0 & -1/6 & -1/2
\end{bmatrix}
$$
so the inverse is
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1/3 & 0 \\
0 & -1/6 & -1/2
\end{bmatrix}
$$
and the matrix with respect to the canonical bases is
$$
\begin{bmatrix}
2a & -4/3 & -2 \\
1 & 1/6 & 1/2
\end{bmatrix}
$$
so the equation becomes
$$
\begin{bmatrix}
2a & -4/3 & -2 \\
1 & 1/6 & 1/2
\end{bmatrix}
\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}
=
\begin{bmatrix} 1 \\ -5/2 \end{bmatrix}
$$
or, doing the computation in the left-hand side,
$$
\begin{bmatrix} 4a-2 \\ 5/2 \end{bmatrix} = 
\begin{bmatrix} 1 \\ -5/2 \end{bmatrix}
$$
Note that I get essentially the same result as you. There is no solution.
