Find the inverse of the following matrix:
[ 2 5 3
[3 1 2 ]
[1 2 1]

To find the inverse of a 3x3 matrix, we can use the formula:

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

where \( \text{det}(A) \) is the determinant of the matrix \( A \) and \( \text{adj}(A) \) is the adjugate of \( A \).

Given the matrix:
\[
A = \begin{bmatrix} 2 & 5 & 3 \\ 3 & 1 & 2 \\ 1 & 2 & 1 \end{bmatrix},
\]

### Step 1: Find the Determinant of \( A \)

The determinant of a 3x3 matrix \( A \) is calculated as:
\[
\text{det}(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}),
\]

Substitute the values:
\[
\text{det}(A) = 2 \times (1 \times 1 - 2 \times 2) - 5 \times (3 \times 1 - 2 \times 1) + 3 \times (3 \times 2 - 1 \times 1).
\]

\[
\text{det}(A) = 2 \times (1 - 4) - 5 \times (3 - 2) + 3 \times (6 - 1),
\]

\[
\text{det}(A) = 2 \times (-3) - 5 \times 1 + 3 \times 5,
\]

\[
\text{det}(A) = -6 - 5 + 15,
\]

\[
\text{det}(A) = 4.
\]

### Step 2: Find the Adjugate of \( A \)

The adjugate of \( A \), denoted \( \text{adj}(A) \), is the transpose of the cofactor matrix of \( A \).

1. **Cofactor of \( a_{11} \):**
\[
C_{11} = \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = (1 \times 1) - (2 \times 2) = 1 - 4 = -3.
\]

2. **Cofactor of \( a_{12} \):**
\[
C_{12} = -\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} = -((3 \times 1) - (2 \times 1)) = -(3 - 2) = -1.
\]

3. **Cofactor of \( a_{13} \):**
\[
C_{13} = \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = (3 \times 2) - (1 \times 1) = 6 - 1 = 5.
\]

4. **Cofactor of \( a_{21} \):**
\[
C_{21} = -\begin{vmatrix} 5 & 3 \\ 2 & 1 \end{vmatrix} = -((5 \times 1) - (3 \times 2)) = -(5 - 6) = 1.
\]

5. **Cofactor of \( a_{22} \):**
\[
C_{22} = \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = (2 \times 1) - (3 \times 1) = 2 - 3 = -1.
\]

6. **Cofactor of \( a_{23} \):**
\[
C_{23} = -\begin{vmatrix} 2 & 5 \\ 1 & 2 \end{vmatrix} = -((2 \times 2) - (5 \times 1)) = -(4 - 5) = 1.
\]

7. **Cofactor of \( a_{31} \):**
\[
C_{31} = \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = (5 \times 2) - (3 \times 1) = 10 - 3 = 7.
\]

8. **Cofactor of \( a_{32} \):**
\[
C_{32} = -\begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = -((2 \times 2) - (3 \times 1)) = -(4 - 3) = -1.
\]

9. **Cofactor of \( a_{33} \):**
\[
C_{33} = \begin{vmatrix} 2 & 5 \\ 3 & 1 \end{vmatrix} = (2 \times 1) - (5 \times 3) = 2 - 15 = -13.
\]

Thus, the cofactor matrix is:
\[
\begin{bmatrix} -3 & -1 & 5 \\ 1 & -1 & 1 \\ 7 & -1 & -13 \end{bmatrix}.
\]

Now, take the transpose to find the adjugate:
\[
\text{adj}(A) = \begin{bmatrix} -3 & 1 & 7 \\ -1 & -1 & -1 \\ 5 & 1 & -13 \end{bmatrix}.
\]

### Step 3: Find the Inverse

\[
A^{-1} = \frac{1}{4} \begin{bmatrix} -3 & 1 & 7 \\ -1 & -1 & -1 \\ 5 & 1 & -13 \end{bmatrix}.
\]

\[
A^{-1} = \begin{bmatrix} -\frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ \frac{5}{4} & \frac{1}{4} & -\frac{13}{4} \end{bmatrix}.
\]

This is the inverse of the matrix \( A \).

Would you like more details on this solution or have any questions?

Here are 5 related questions:
1. How do you find the inverse of a matrix using row reduction?
2. When does a matrix not have an inverse?
3. What is the geometric interpretation of the inverse of a matrix?
4. How does the inverse of a matrix relate to solving systems of linear equations?
5. What are the properties of an invertible matrix?

**Tip:** A matrix has an inverse if and only if its determinant is non-zero.

Find the inverse of the following matrix: [2 5 3, 3 1 2, 1 2 1]

Discover how to calculate the inverse of a 3x3 matrix [2 5 3, 3 1 2, 1 2 1] using the determinant and adjugate methods. This detailed solution covers key concepts in linear algebra, including matrix inversion and determinant calculation, suitable for advanced high school or college students.

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