## Subject 1.

$$
\text { Determine the matrix } A \in M_{3}(C) \text {, if } A^{*}=\left(\begin{array}{ccc}
-5 & 7 & 1 \\
1 & -5 & 7 \\
7 & 1 & -5
\end{array}\right) \text {. }
$$

G.M. $12 / 2015$
## Solution:

Write the relation $A \cdot A^{-1}=A^{-1} \cdot A=I_{3}$, det $A \neq 0$, and $A^{-1}=\frac{1}{\operatorname{det} A} \cdot A^{*}$ $\qquad$ (1p).

Justify that $A^{*}$ is invertible, det $A^{*}=324$, and calculate its inverse $\left(A^{*}\right)^{-1}=\frac{1}{18}\left(\begin{array}{lll}1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1\end{array}\right) \ldots \ldots$. (3p).

Show that $A=(\operatorname{det} A) \cdot\left(A^{*}\right)^{-1}$ (1p).

Show that $\operatorname{det} A^{*}=(\operatorname{det} A)^{2}, \operatorname{det} A= \pm 18$ (1p).

Finalize $A= \pm\left(\begin{array}{lll}1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1\end{array}\right)$