 (3 points) Find the inverse of the following matrix A.
A =


1 2 2
1 1 2
2 4 3



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

Given matrix \( A \):

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

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

The determinant of a 3x3 matrix is calculated as:

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

For matrix \( A \), with elements:

\[
\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix} = 
\begin{pmatrix}
1 & 2 & 2 \\
1 & 1 & 2 \\
2 & 4 & 3
\end{pmatrix}
\]

Substitute the values into the determinant formula:

\[
\text{det}(A) = 1 \left( 1(3) - 2(4) \right) - 2 \left( 1(3) - 2(2) \right) + 2 \left( 1(4) - 1(2) \right)
\]
\[
\text{det}(A) = 1(3 - 8) - 2(3 - 4) + 2(4 - 2)
\]
\[
\text{det}(A) = 1(-5) - 2(-1) + 2(2)
\]
\[
\text{det}(A) = -5 + 2 + 4 = 1
\]

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

### Step 2: Find the matrix of minors

The matrix of minors is formed by calculating the determinant of each 2x2 minor matrix for each element of \( A \).

For example, the minor for element \( a_{11} \) (first row, first column) is:

\[
\text{Minor}(a_{11}) = \begin{vmatrix} 1 & 2 \\ 4 & 3 \end{vmatrix} = (1)(3) - (2)(4) = 3 - 8 = -5
\]

Repeat this for each element of matrix \( A \):

\[
\text{Matrix of Minors} =
\begin{pmatrix}
-5 & -1 & 2 \\
-2 & -1 & 2 \\
0 & 0 & -1
\end{pmatrix}
\]

### Step 3: Find the matrix of cofactors

To form the matrix of cofactors, apply a sign change based on the position of each element using the checkerboard pattern of signs (+, -, +; -, +, -; +, -, +):

\[
\text{Matrix of Cofactors} =
\begin{pmatrix}
-5 & 1 & 2 \\
2 & -1 & -2 \\
0 & 0 & -1
\end{pmatrix}
\]

### Step 4: Find the adjugate matrix

The adjugate matrix is the transpose of the matrix of cofactors:

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

### Step 5: Multiply by \( 1/\text{det}(A) \)

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

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

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 further explanation on any of these steps?

### Related Questions:
1. What are the conditions for a matrix to have an inverse?
2. How is the determinant of a 3x3 matrix related to its invertibility?
3. Can a matrix with a zero determinant have an inverse? Why or why not?
4. How would the inverse change if the determinant were different?
5. How does the adjugate matrix help in finding the inverse?

### Tip:
Always verify the determinant before attempting to find the inverse of a matrix. If the determinant is zero, the matrix does not have an inverse.

Find the inverse of the following matrix A.
A = 
\begin{pmatrix}
1 & 2 & 2 \\
1 & 1 & 2 \\
2 & 4 & 3
\end{pmatrix}

This detailed solution explains how to find the inverse of a 3x3 matrix A, with a step-by-step breakdown including calculating the determinant, matrix of minors, cofactors, adjugate matrix, and final inverse. Suitable for undergraduate students studying linear algebra.

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