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To find the inverse of the given matrix \( A \), we will follow these steps:

Given the matrix \( A \):

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

### Steps to find the inverse:

#### Step 1: Calculate the determinant of \( A \)
The formula for the determinant of a 3x3 matrix is:
\[
\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
\]
For the matrix \( A = \begin{pmatrix} 
1 & 2 & 2 \\
1 & 1 & 2 \\
2 & 4 & 3 
\end{pmatrix} \), 

the determinant is:
\[
\text{det}(A) = 1((1)(3) - (2)(4)) - 2((1)(3) - (2)(2)) + 2((1)(4) - (1)(2))
\]
\[
= 1(3 - 8) - 2(3 - 4) + 2(4 - 2)
\]
\[
= 1(-5) - 2(-1) + 2(2)
\]
\[
= -5 + 2 + 4 = 1
\]

Since the determinant is \( 1 \), the matrix \( A \) is invertible.

#### Step 2: Find the cofactor matrix
To find the inverse of \( A \), we need to compute the cofactor matrix. We compute the minors for each element of the matrix and apply the necessary signs.

The cofactor matrix \( C \) of \( A \) is:

\[
C = \begin{pmatrix} 
\text{det} \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix} & -\text{det} \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} & \text{det} \begin{pmatrix} 1 & 1 \\ 2 & 4 \end{pmatrix} \\
-\text{det} \begin{pmatrix} 2 & 2 \\ 4 & 3 \end{pmatrix} & \text{det} \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} & -\text{det} \begin{pmatrix} 1 & 1 \\ 2 & 4 \end{pmatrix} \\
\text{det} \begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix} & -\text{det} \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix} & \text{det} \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}
\end{pmatrix}
\]

Computing the minors:

- \( \text{det} \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix} = (1)(3) - (2)(4) = 3 - 8 = -5 \)
- \( \text{det} \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} = (1)(3) - (2)(2) = 3 - 4 = -1 \)
- \( \text{det} \begin{pmatrix} 1 & 1 \\ 2 & 4 \end{pmatrix} = (1)(4) - (1)(2) = 4 - 2 = 2 \)
- \( \text{det} \begin{pmatrix} 2 & 2 \\ 4 & 3 \end{pmatrix} = (2)(3) - (2)(4) = 6 - 8 = -2 \)
- \( \text{det} \begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix} = (2)(1) - (2)(1) = 2 - 2 = 0 \)
- \( \text{det} \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = (1)(2) - (1)(1) = 2 - 1 = 1 \)

Thus, the cofactor matrix \( C \) is:

\[
C = \begin{pmatrix} 
-5 & 1 & 2 \\
2 & -1 & -2 \\
0 & 0 & 1
\end{pmatrix}
\]

#### Step 3: Transpose the cofactor matrix
The adjugate matrix (transpose of the cofactor matrix) is:

\[
C^T = \begin{pmatrix} 
-5 & 2 & 0 \\
1 & -1 & 0 \\
2 & -2 & 1
\end{pmatrix}
\]

#### Step 4: Multiply by the inverse of the determinant
Since the determinant is \( 1 \), the inverse of matrix \( A \) is simply the adjugate matrix:

\[
A^{-1} = \frac{1}{\text{det}(A)} C^T = C^T
\]

Thus, the inverse of matrix \( A \) is:

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

Would you like more details or explanations on any of the steps? Here are 5 related questions to expand your understanding:

1. What is the determinant of a 2x2 matrix, and how does it generalize to higher dimensions?
2. Why is the determinant an essential step in finding the inverse of a matrix?
3. How does the cofactor matrix differ from the adjugate matrix?
4. Can you explain what it means for a matrix to be invertible?
5. What are some real-world applications of finding the inverse of a matrix?

**Tip:** Always check the determinant before trying to find the inverse of a matrix. If the determinant is zero, the matrix is non-invertible (singular).

Find the inverse of the following matrix A:
A = 
1 2 2
1 1 2
2 4 3

Learn how to find the inverse of the 3x3 matrix A, with a detailed solution involving calculating the determinant, cofactor matrix, and adjugate. This problem is suitable for undergraduate students studying linear algebra.

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