1.28. There are linear transformations $A x=\left\{x_{2}-x_{3}\right.$, $\left.x_{1}, x_{1}+x_{3}\right\}$ and $B x=\left\{x_{2}, 2 x_{2}, x_{1}\right\}$ of the vector $x=\left\{x_{1}, x_{2}, x_{3}\right\}$. Find the mapping $B \cdot(2 A-B) x$.
Solution. Let's write down the matrices of the linear mappings. For this, we will write the given mappings in matrix form:

$$
\left(\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right) \cdot\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right)=\left(\begin{array}{c}
x_{2}-x_{3} \\
x_{1} \\
x_{1}+x_{3}
\end{array}\right) ;\left(\begin{array}{lll}
b_{11} & b_{12} & b_{13} \\
b_{21} & b_{22} & b_{23} \\
b_{31} & b_{32} & b_{33}
\end{array}\right) \cdot\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right)=\left(\begin{array}{c}
x_{2} \\
2 x_{2} \\
x_{1}
\end{array}\right) \cdot
$$

Multiplying the matrices in the left parts of the written matrix equations, we get

$$
\left(\begin{array}{c}
a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3} \\
a_{21} x_{1}+a_{22} x_{2}+a_{23} x_{3} \\
a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}
\end{array}\right)=\left(\begin{array}{c}
x_{2}-x_{3} \\
x_{1} \\
x_{1}+x_{3}
\end{array}\right) ;\left(\begin{array}{c}
b_{11} x_{1}+b_{12} x_{2}+b_{13} x_{3} \\
b_{21} x_{1}+b_{22} x_{2}+b_{23} x_{3} \\
b_{31} x_{1}+b_{32} x_{2}+b_{33} x_{3}
\end{array}\right)=\left(\begin{array}{c}
x_{2} \\
2 x_{2} \\
x_{1}
\end{array}\right)
$$

From this, it follows that

$$
\begin{aligned}
& a_{11}=0, \quad a_{12}=1, a_{13}=-1 \\
& a_{21}=1, \quad a_{22}=0, \quad a_{23}=0 \\
& a_{31}=1, a_{32}=0, a_{33}=1 \\
& b_{11}=0, \quad b_{12}=1, \quad b_{13}=0 \\
& b_{21}=0, b_{22}=2, b_{23}=0 \\
& b_{31}=1, b_{32}=0, b_{33}=0
\end{aligned}
$$

Thus, the matrices $A$ and $B$ are

$$
A=\left(\begin{array}{ccc}
0 & 1 & -1 \\
1 & 0 & 0 \\
1 & 0 & 1
\end{array}\right), B=\left(\begin{array}{lll}
0 & 1 & 0 \\
0 & 2 & 0 \\
1 & 0 & 0
\end{array}\right)
$$

Taking into account the properties of the linear transformation, we find the matrix of the desired linear mapping:

$$
B \cdot(2 A-B)=2 B \cdot A-B \cdot B=
$$

$=2 \cdot\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 2 & 0 \\ 1 & 0 & 0\end{array}\right) \cdot\left(\begin{array}{ccc}0 & 1 & -1 \\ 1 & 0 & 0 \\ 1 & 0 & 1\end{array}\right)-\left(\begin{array}{ccc}0 & 1 & 0 \\ 0 & 2 & 0 \\ 1 & 0 & 0\end{array}\right) \cdot\left(\begin{array}{ccc}0 & 1 & 0 \\ 0 & 2 & 0 \\ 1 & 0 & 0\end{array}\right)=\left(\begin{array}{ccc}2 & -2 & 0 \\ 0 & -4 & 0 \\ 1 & 1 & -2\end{array}\right)$