### 1.26. Find the coordinates of the vector

$$
X=\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right)
$$

if the linear transformation that maps it to the vector

$$
Y=\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)
$$

is given by the system of equations

$$
\left\{\begin{array}{l}
x_{1}+2 x_{2}+3 x_{3}=1 \\
-x_{1}+3 x_{2}+4 x_{3}=2 \\
2 x_{1}-3 x_{2}+5 x_{3}=3
\end{array}\right.
$$
Solution. We find the inverse matrix for the main matrix of the system

$$
A=\left(\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 3 & 4 \\
2 & -3 & 5
\end{array}\right)
$$

We compute the determinant of this matrix to ensure the existence of the inverse matrix. We have

![](https://cdn.mathpix.com/cropped/2024_05_22_f5a9bb7960904462604bg-064.jpg?height=138&width=917&top_left_y=1474&top_left_x=177)

$$
\begin{aligned}
& A^{-1}=\frac{1}{44}\left(\begin{array}{ccc}
27 & -19 & -1 \\
13 & -1 & -7 \\
-3 & 7 & 5
\end{array}\right)
\end{aligned}
$$

From this,

$$
X=A^{-1} \cdot B=\left(\begin{array}{c}
-\frac{7}{22} \\
-\frac{5}{22} \\
\frac{13}{22}
\end{array}\right)
$$

The obtained solution gives the coordinates of the desired vector.