Example. Given a third-order square matrix

$$
C=\left(\begin{array}{rrr}
1 & 2 & 1 \\
3 & -5 & 3 \\
2 & 7 & -1
\end{array}\right)
$$

Determine the existence and find the inverse matrix $C^{-1}$.
Solution.

1. We compute the determinant of matrix $\operatorname{det} C$:

$$
\operatorname{det} C=\left|\begin{array}{rrr}
1 & 2 & 1 \\
3 & -5 & 3 \\
2 & 7 & -1
\end{array}\right|=1 \cdot(-16)-2 \cdot(-9)+1 \cdot 31=33
$$

Since $\operatorname{det} C \neq 0$, the matrix $C$ has an inverse.

2. We form the matrix of cofactors

$$
\widetilde{C}=\left(\begin{array}{rrr}
-16 & 9 & 31 \\
9 & -3 & -3 \\
11 & 0 & -11
\end{array}\right)
$$

3. We transpose the matrix $\tilde{C}$

$$
\tilde{C}^{\mathrm{T}}=\left(\begin{array}{rrr}
-16 & 9 & 11 \\
9 & -3 & 0 \\
31 & -3 & -11
\end{array}\right)
$$

4. By dividing the matrix $\tilde{C}^{\mathrm{T}}$ by the determinant, we obtain the desired inverse matrix

$$
C^{-1}=\frac{1}{33} \cdot\left(\begin{array}{rrr}
-16 & 9 & 11 \\
9 & -3 & 0 \\
31 & -3 & -11
\end{array}\right)
$$

5. We verify

$$
C \cdot C^{-1}=\frac{1}{33} \cdot\left(\begin{array}{rrr}
-16 & 9 & 11 \\
9 & -3 & 0 \\
31 & -3 & -11
\end{array}\right) \cdot\left(\begin{array}{rrr}
1 & 2 & 1 \\
3 & -5 & 3 \\
2 & 7 & -1
\end{array}\right)=\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right)
$$

Answer. The matrix inverse to matrix $C$ is

$$
C^{-1}=\frac{1}{33} \cdot\left(\begin{array}{rrr}
-16 & 9 & 11 \\
9 & -3 & 0 \\
31 & -3 & -11
\end{array}\right)
$$

Conditions of the problems. Find the matrices inverse to the given ones.

1. $\left(\begin{array}{lll}2 & 3 & 1 \\ 3 & 4 & 2 \\ 1 & 1 & 2\end{array}\right) \cdot 2$.
2. $\left(\begin{array}{lll}4 & 3 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
3. $\left(\begin{array}{lll}0 & 0 & 1 \\ 1 & 5 & 1 \\ 1 & 4 & 2\end{array}\right)$
4. $\left(\begin{array}{lll}1 & 2 & 1 \\ 0 & 3 & 1 \\ 0 & 2 & 1\end{array}\right)$
5. $\left(\begin{array}{lll}1 & 1 & 2 \\ 0 & 2 & 3 \\ 0 & 1 & 2\end{array}\right)$
6. $\left(\begin{array}{lll}2 & 0 & 5 \\ 3 & 1 & 5 \\ 1 & 0 & 2\end{array}\right)$
7. $\left(\begin{array}{ccc}1 & 0 & 1 \\ 2 & 2 & 3 \\ 3 & 0 & 4\end{array}\right)$
8. $\left(\begin{array}{lll}2 & 1 & 0 \\ 3 & 2 & 0 \\ 3 & 2 & 1\end{array}\right)$
9. $\left(\begin{array}{lll}1 & 3 & 0 \\ 1 & 2 & 0 \\ 2 & 1 & 1\end{array}\right)$

Answers.

1. $\left(\begin{array}{rrr}-6 & 5 & -2 \\ 3 & -3 & 1 \\ 1 & 1 & 2\end{array}\right) \cdot 2 \cdot\left(\begin{array}{rrr}1 & -3 & 0 \\ -1 & 4 & 0 \\ 0 & 0 & 1\end{array}\right) \cdot 3 \cdot\left(\begin{array}{rrr}-6 & -4 & -5 \\ 1 & 1 & -1 \\ 1 & 0 & 0\end{array}\right)$.
2. $\left(\begin{array}{rrr}1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & -2 & 3\end{array}\right)$.
3. $\left(\begin{array}{rrr}1 & 0 & -1 \\ 0 & 2 & -3 \\ 0 & -1 & 2\end{array}\right)$.
4. $\left(\begin{array}{rrr}-2 & 0 & 5 \\ 1 & 1 & -5 \\ 1 & 0 & -2\end{array}\right)$.
5. $\left(\begin{array}{rrr}4 & 0 & -1 \\ 1 & 1 & -1 \\ -3 & 0 & 1\end{array}\right)$.
6. $\left(\begin{array}{rrr}2 & -1 & 0 \\ -3 & 2 & 0 \\ 0 & -1 & 1\end{array}\right)$.
7. $\left(\begin{array}{rrr}-2 & 3 & 0 \\ 1 & -1 & 0 \\ 3 & -5 & 1\end{array}\right)$.

## 2.3. The Concept of a Linear Space

Problem Statement. Does the given set $X$, in which the "sum" $a \oplus b$ of any two elements $a$ and $b$ and the "product" $\alpha \odot a$ of any element $a$ by any number $\alpha$ are defined, form a linear space?

Plan of Solution. Based on the definition of a linear space, we check the following conditions.

1. Are the introduced operations of addition and multiplication by a number closed in $X$, i.e., is it true that $\forall a, b \in X$ and $\forall \alpha \in \mathbb{R}$

$$
a \oplus b \in X, \quad \alpha \odot a \in X ?
$$

If not, then the set $X$ is not a linear space; if yes, we continue the check.

2. We find the zero element $\theta \in X$ such that $\forall a \in X$

$$
a \oplus \theta=a
$$

If such an element does not exist, then the set $X$ is not a linear space; if it exists, we continue the check.

3. For each element $a \in X$, we determine the opposite element $-a \in X$ such that

$$
a \oplus-a=\theta
$$

If such an element does not exist, then the set $X$ is not a linear space; if it exists, we continue the check.

4. We check the fulfillment of the remaining axioms of a linear space, i.e., $\forall a, b, c \in X$ and $\forall \alpha, \beta \in \mathbb{R}$:

$a \oplus b=b \oplus a$

$(a \oplus b) \oplus c=a \oplus(b \oplus c)$

$\alpha \odot(\beta \odot a)=(\alpha \cdot \beta) \odot a$

$1 \odot a=a$

$(\alpha+\beta) \odot a=\alpha \odot a \oplus \beta \odot a ;$

$\alpha \odot(a \oplus b)=\alpha \odot a \oplus \alpha \odot b$.

If any of the axioms are violated, then the set $X$ is not a linear space. If all axioms are satisfied, then the set $X$ is a linear space.