1.29. Let the matrix $A=\left(\begin{array}{ccc}1 & 1 & 2 \\ 3 & -1 & 0 \\ 1 & 1 & -2\end{array}\right)$ be used to 

transform the vector $x$ in the basis $\left\{e_{1}, e_{2}, e_{3}\right\}$. Determine the transformation matrix $A^{\prime}$ in the basis $\left\{e_{1}^{\prime}, e_{2}^{\prime}, e_{3}^{\prime}\right\}$, if

$$
e_{1}^{\prime}=e_{1}-e_{2}+e_{3}, \quad e_{2}^{\prime}=e_{1}+e_{2}-2 e_{3}, \quad e_{3}^{\prime}=-e_{1}+2 e_{2}+e_{3} .
$$
Solution. Let's write down the transformation matrix from the old basis to the new one:

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

Find the inverse matrix $B^{-1}$, first ensuring that it is not singular:

$$
\begin{aligned}
& \operatorname{det} B=\left|\begin{array}{ccc}
0 & -1 & 1 \\
-1 & 1 & -2 \\
-1 & 2 & 1
\end{array}\right|=1 \neq 0, B^{-1}=\left(\begin{array}{ccc}
5 & 3 & 1 \\
3 & 2 & 1 \\
-1 & -1 & 0
\end{array}\right) \\
& B^{-1} \cdot A=\left(\begin{array}{ccc}
5 & 3 & 1 \\
3 & 2 & 1 \\
-1 & -1 & 0
\end{array}\right) \cdot\left(\begin{array}{ccc}
1 & 1 & 2 \\
3 & -1 & 0 \\
1 & 1 & -2
\end{array}\right)=\left(\begin{array}{ccc}
15 & -2 & 8 \\
8 & -1 & 4 \\
-4 & 1 & -2
\end{array}\right) .
\end{aligned}
$$

Find the desired transformation matrix using the formula

$$
A^{\prime}=B^{-1} \cdot A \cdot B=\left(\begin{array}{ccc}
15 & -2 & 8 \\
8 & -1 & 4 \\
-4 & 1 & -2
\end{array}\right) \cdot\left(\begin{array}{ccc}
1 & -1 & 1 \\
-1 & 1 & -2 \\
-1 & 2 & 1
\end{array}\right)=\left(\begin{array}{ccc}
9 & -1 & 27 \\
5 & -1 & 14 \\
-3 & 1 & -8
\end{array}\right)
$$