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The question asks to find the inverse of the given matrix \( A \), which is:

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

### (a) Finding the Inverse of Matrix \( A \)

To find the inverse of a 3x3 matrix \( A \), we can use the following method:

1. **Find the determinant of \( A \)**:
   
   The determinant of \( A \) is calculated as:

   \[
   \text{det}(A) = 1 \cdot \begin{vmatrix} 1 & 2 \\ 4 & 3 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} + 2 \cdot \begin{vmatrix} 1 & 1 \\ 2 & 4 \end{vmatrix}
   \]
   
   Calculate each minor determinant:

   \[
   \begin{vmatrix} 1 & 2 \\ 4 & 3 \end{vmatrix} = (1)(3) - (4)(2) = 3 - 8 = -5
   \]
   \[
   \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = (1)(3) - (2)(2) = 3 - 4 = -1
   \]
   \[
   \begin{vmatrix} 1 & 1 \\ 2 & 4 \end{vmatrix} = (1)(4) - (1)(2) = 4 - 2 = 2
   \]

   Now substitute these values back into the formula for the determinant:

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

   So, the determinant of \( A \) is 1.

2. **Find the adjugate (adjoint) of \( A \)**:

   The adjugate of a matrix is the transpose of its cofactor matrix. For a 3x3 matrix, each element \( C_{ij} \) of the cofactor matrix is found by calculating the determinant of the 2x2 submatrix that remains after removing the \( i \)-th row and \( j \)-th column, and then multiplying by \( (-1)^{i+j} \).

   We already calculated the necessary minors, so the cofactor matrix is:

   \[
   \text{Cofactor}(A) = \begin{pmatrix} -5 & 1 & 0 \\ 2 & 1 & -2 \\ 1 & -4 & -1 \end{pmatrix}
   \]

   Taking the transpose of the cofactor matrix to get the adjugate matrix:

   \[
   \text{Adj}(A) = \begin{pmatrix} -5 & 2 & 1 \\ 1 & 1 & -4 \\ 0 & -2 & -1 \end{pmatrix}
   \]

3. **Find the inverse**:

   Since \( \text{det}(A) = 1 \), the inverse of \( A \) is simply the adjugate matrix divided by the determinant:

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

   So,

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

### (b) Solving \( Ax = b \)

The problem also asks to solve the system \( Ax = b \), where:

\[
x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \quad b = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}
\]

Using the inverse of \( A \), the solution is given by:

\[
x = A^{-1}b
\]

Let's perform the multiplication:

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

Multiplying:

\[
x_1 = (-5)(1) + (2)(0) + (1)(-1) = -5 + 0 - 1 = -6
\]
\[
x_2 = (1)(1) + (1)(0) + (-4)(-1) = 1 + 0 + 4 = 5
\]
\[
x_3 = (0)(1) + (-2)(0) + (-1)(-1) = 0 + 0 + 1 = 1
\]

So, the solution is:

\[
x = \begin{pmatrix} -6 \\ 5 \\ 1 \end{pmatrix}
\]

---

Would you like more details on any step or further clarifications?

Here are some additional related questions you might find helpful:

1. How can we generalize the method for finding the inverse of a matrix to larger matrices?
2. What are some situations where a matrix might not have an inverse?
3. How can you use matrix inverses in systems of linear equations with more than three variables?
4. What is the role of the determinant in determining whether a matrix is invertible?
5. What are alternative methods to solve a system of equations, like Gaussian elimination?

**Tip:** Always check if a matrix is invertible (i.e., non-zero determinant) before attempting to find its inverse!

Find the inverse of the matrix A and solve the system Ax = b, where A is a 3x3 matrix and b is a column vector.

Learn how to find the inverse of a 3x3 matrix and solve a system of linear equations using matrix inversion. This problem involves determinant calculation, cofactor matrices, and solving Ax = b. Suitable for advanced high school or undergraduate students studying linear algebra.

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