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To find the inverse of the given 3x3 matrix, we will use the formula for the inverse of a square matrix \( A \), where:

\[
A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)
\]

The matrix provided is:

\[
A = \begin{pmatrix}
-1 & 5 & -2 \\
1 & -4 & 2 \\
1 & -3 & 0
\end{pmatrix}
\]

### Step-by-step solution:

#### Step 1: Find the determinant of matrix \( A \).

The determinant \( \det(A) \) of a 3x3 matrix \( A \) is given by:

\[
\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
\]

Where:
\[
A = \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}
\]

For the given matrix \( A \):

\[
\det(A) = (-1) \left( (-4)(0) - (2)(-3) \right) - 5 \left( (1)(0) - (2)(1) \right) + (-2) \left( (1)(-3) - (-4)(1) \right)
\]

Let’s compute each term:

\[
\det(A) = (-1) \left( 0 - (-6) \right) - 5 \left( 0 - 2 \right) + (-2) \left( (-3) - (-4) \right)
\]
\[
\det(A) = (-1)(6) - 5(-2) + (-2)(1)
\]
\[
\det(A) = -6 + 10 - 2 = 2
\]

#### Step 2: Find the adjugate matrix \( \text{adj}(A) \).

The adjugate matrix is the transpose of the cofactor matrix. To find the cofactors, we compute the determinant of the 2x2 submatrices for each element.

##### Cofactors:

1. \( C_{11} = \det \begin{pmatrix} -4 & 2 \\ -3 & 0 \end{pmatrix} = (-4)(0) - (2)(-3) = 6 \)
2. \( C_{12} = -\det \begin{pmatrix} 1 & 2 \\ 1 & 0 \end{pmatrix} = -(1 \cdot 0 - 2 \cdot 1) = 2 \)
3. \( C_{13} = \det \begin{pmatrix} 1 & -4 \\ 1 & -3 \end{pmatrix} = 1 \cdot (-3) - (-4) \cdot 1 = 1 \)
4. \( C_{21} = -\det \begin{pmatrix} 5 & -2 \\ -3 & 0 \end{pmatrix} = -(5 \cdot 0 - (-2) \cdot (-3)) = -6 \)
5. \( C_{22} = \det \begin{pmatrix} -1 & -2 \\ 1 & 0 \end{pmatrix} = (-1)(0) - (-2)(1) = 2 \)
6. \( C_{23} = -\det \begin{pmatrix} -1 & 5 \\ 1 & 1 \end{pmatrix} = -((-1)(1) - (5)(1)) = -(-1 - 5) = 6 \)
7. \( C_{31} = \det \begin{pmatrix} 5 & -2 \\ -4 & 2 \end{pmatrix} = (5)(2) - (-2)(-4) = 10 - 8 = 2 \)
8. \( C_{32} = -\det \begin{pmatrix} -1 & -2 \\ 1 & 2 \end{pmatrix} = -((-1)(2) - (-2)(1)) = 0 \)
9. \( C_{33} = \det \begin{pmatrix} -1 & 5 \\ 1 & -4 \end{pmatrix} = (-1)(-4) - (5)(1) = 4 - 5 = -1 \)

Now the cofactor matrix is:

\[
\text{Cofactor matrix} = \begin{pmatrix}
6 & 2 & 1 \\
-6 & 2 & 6 \\
2 & 0 & -1
\end{pmatrix}
\]

The adjugate matrix is the transpose of the cofactor matrix:

\[
\text{adj}(A) = \begin{pmatrix}
6 & -6 & 2 \\
2 & 2 & 0 \\
1 & 6 & -1
\end{pmatrix}
\]

#### Step 3: Find the inverse matrix.

Now that we have \( \det(A) = 2 \) and the adjugate matrix \( \text{adj}(A) \), the inverse matrix is:

\[
A^{-1} = \frac{1}{2} \cdot \begin{pmatrix}
6 & -6 & 2 \\
2 & 2 & 0 \\
1 & 6 & -1
\end{pmatrix}
\]

Thus, the inverse matrix is:

\[
A^{-1} = \begin{pmatrix}
3 & -3 & 1 \\
1 & 1 & 0 \\
\frac{1}{2} & 3 & -\frac{1}{2}
\end{pmatrix}
\]

---

Would you like further clarifications or details? Here are 5 related questions you might find useful:

1. How is the determinant of a matrix calculated in general?
2. Why is the determinant important for finding the inverse of a matrix?
3. What are other methods for finding an inverse, such as Gaussian elimination?
4. How do the cofactors contribute to the adjugate matrix?
5. What happens if the determinant of a matrix is zero?

**Tip:** If the determinant of a matrix is 0, it means the matrix is **singular** and does not have an inverse!

Learn how to find the inverse of a 3x3 matrix with a step-by-step guide. This solution covers key linear algebra concepts, including calculating the determinant, finding the adjugate, and applying the matrix inversion formula. Suitable for high school and university students.

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