Using the inverse matrix method, solve the following set of simultaneous equations:
2y + x + 2z = 10
x − z + y = 12
2z + x − 2y = 6

To solve the given system of simultaneous equations using the inverse matrix method, we first need to express the system in matrix form \( AX = B \), where:

\[
A = \begin{pmatrix} 
1 & 2 & 2 \\
1 & 1 & -1 \\
1 & -2 & 2 
\end{pmatrix}, \quad
X = \begin{pmatrix} 
x \\
y \\
z 
\end{pmatrix}, \quad
B = \begin{pmatrix} 
10 \\
12 \\
6 
\end{pmatrix}
\]

### Step 1: Matrix Form
We can rewrite the system of equations as follows:

\[
\begin{pmatrix} 
1 & 2 & 2 \\
1 & 1 & -1 \\
1 & -2 & 2 
\end{pmatrix} 
\begin{pmatrix} 
x \\
y \\
z 
\end{pmatrix}
=
\begin{pmatrix} 
10 \\
12 \\
6 
\end{pmatrix}
\]

### Step 2: Finding the Inverse of Matrix \( A \)
We need to find the inverse of the coefficient matrix \( A \), denoted \( A^{-1} \). The inverse of a 3x3 matrix can be found using the formula involving determinants and cofactors, but for the sake of simplicity, let's compute it directly:

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

First, let's calculate the determinant of \( A \):

\[
\text{det}(A) = 1 \cdot \left( (1)(2) - (-1)(-2) \right) - 2 \cdot \left( (1)(2) - (-1)(1) \right) + 2 \cdot \left( (1)(-2) - (1)(1) \right)
\]

\[
\text{det}(A) = 1 \cdot (2 - 2) - 2 \cdot (2 + 1) + 2 \cdot (-2 - 1)
\]
\[
\text{det}(A) = 0 - 6 + (-6) = -12
\]

Now, we find the adjugate matrix of \( A \), then multiply by \( 1/\text{det}(A) \). After computing:

\[
A^{-1} = \frac{1}{-12} \cdot \begin{pmatrix}
6 & -8 & -6 \\
4 & 4 & -4 \\
-2 & 4 & 6
\end{pmatrix}
\]

Thus,

\[
A^{-1} = \begin{pmatrix}
-\frac{1}{2} & \frac{2}{3} & \frac{1}{2} \\
-\frac{1}{3} & -\frac{1}{3} & \frac{1}{3} \\
\frac{1}{6} & -\frac{1}{3} & -\frac{1}{2}
\end{pmatrix}
\]

### Step 3: Multiply \( A^{-1} \) by \( B \)
Now, multiply the inverse matrix \( A^{-1} \) by the column vector \( B \):

\[
X = A^{-1}B = \begin{pmatrix}
-\frac{1}{2} & \frac{2}{3} & \frac{1}{2} \\
-\frac{1}{3} & -\frac{1}{3} & \frac{1}{3} \\
\frac{1}{6} & -\frac{1}{3} & -\frac{1}{2}
\end{pmatrix} 
\begin{pmatrix} 
10 \\
12 \\
6 
\end{pmatrix}
\]

\[
X = \begin{pmatrix} 
-\frac{1}{2}(10) + \frac{2}{3}(12) + \frac{1}{2}(6) \\
-\frac{1}{3}(10) - \frac{1}{3}(12) + \frac{1}{3}(6) \\
\frac{1}{6}(10) - \frac{1}{3}(12) - \frac{1}{2}(6)
\end{pmatrix}
\]

After calculation:

\[
X = \begin{pmatrix} 
-5 + 8 + 3 \\
-\frac{10}{3} - 4 + 2 \\
\frac{10}{6} - 4 - 3
\end{pmatrix} 
=
\begin{pmatrix} 
6 \\
-4 \\
-3
\end{pmatrix}
\]

### Step 4: Final Solution
Thus, the solution to the system is:

\[
x = 6, \quad y = -4, \quad z = -3
\]

Let me know if you'd like more details on any step or if you have any questions!

---

Here are five related questions to expand on this solution:
1. How can you verify if the inverse matrix method works for a different system of equations?
2. What happens if the determinant of the matrix \( A \) is zero?
3. How do you find the adjugate matrix in a 3x3 system?
4. Can the inverse matrix method be used for larger systems of equations, such as 4x4?
5. How does the inverse matrix method compare to other methods like Gaussian elimination?

**Tip**: Always check the determinant of the matrix before applying the inverse matrix method. If it's zero, the system has no unique solution.

Using the inverse matrix method, solve the following set of simultaneous equations: 2y + x + 2z = 10, x − z + y = 12, 2z + x − 2y = 6

Learn how to solve a set of simultaneous equations using the inverse matrix method. This detailed step-by-step solution covers matrix operations, determinants, and applying the inverse matrix to solve equations. Suitable for advanced high school students or early college learners.

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