1.27. Given a linear transformation using a system of equations

$$
\left\{\begin{array}{l}
g_{1}=-x_{1}+0 \cdot x_{2}+0 \cdot x_{3} \\
g_{2}=0 \cdot x_{1}+1 \cdot x_{2}+0 \cdot x_{3} \\
g_{3}=0 \cdot x_{1}+0 \cdot x_{2}+1 \cdot x_{3}
\end{array}\right.
$$

Find the transformation matrix, will it be singular?
Solution. With the matrix representation of the transformation, we get:

$$
\left(\begin{array}{l}
g_{1} \\
g_{2} \\
g_{3}
\end{array}\right)=\left(\begin{array}{ccc}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right) \cdot\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right)
$$

Here, the transformation matrix

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

has a non-zero determinant, equal to the product of all elements on the main diagonal \(\operatorname{det} A=(-1) \cdot 1 \cdot 1=-1 \neq 0\) (the mapping is also non-degenerate).