Solve the following system of equations by Cramer’s rule
x + y + z = 5, 3x − 2y + 2z = 3, 2x + 2y − z = 4

To solve the system of equations using **Cramer's Rule**, we follow these steps:

### Given system of equations:
\[
\begin{align*}
1. & \quad x + y + z = 5 \\
2. & \quad 3x - 2y + 2z = 3 \\
3. & \quad 2x + 2y - z = 4
\end{align*}
\]

### Step 1: Write the system in matrix form

This system can be written in matrix form as:
\[
\mathbf{A} \mathbf{x} = \mathbf{b}
\]
where:
\[
\mathbf{A} =
\begin{pmatrix}
1 & 1 & 1 \\
3 & -2 & 2 \\
2 & 2 & -1
\end{pmatrix}
,\quad
\mathbf{x} = 
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}
,\quad
\mathbf{b} =
\begin{pmatrix}
5 \\
3 \\
4
\end{pmatrix}
\]

### Step 2: Find the determinant of matrix \( A \)

We first calculate the determinant of the coefficient matrix \( \mathbf{A} \):
\[
\det(\mathbf{A}) = \begin{vmatrix} 
1 & 1 & 1 \\ 
3 & -2 & 2 \\ 
2 & 2 & -1 
\end{vmatrix}
\]
Using cofactor expansion along the first row:
\[
\det(\mathbf{A}) = 1 \cdot \begin{vmatrix} -2 & 2 \\ 2 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & -2 \\ 2 & 2 \end{vmatrix}
\]

Now, calculate the 2x2 determinants:
\[
\det \begin{vmatrix} -2 & 2 \\ 2 & -1 \end{vmatrix} = (-2)(-1) - (2)(2) = 2 - 4 = -2
\]
\[
\det \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} = (3)(-1) - (2)(2) = -3 - 4 = -7
\]
\[
\det \begin{vmatrix} 3 & -2 \\ 2 & 2 \end{vmatrix} = (3)(2) - (-2)(2) = 6 + 4 = 10
\]

Now substitute back:
\[
\det(\mathbf{A}) = 1(-2) - 1(-7) + 1(10) = -2 + 7 + 10 = 15
\]

### Step 3: Find determinants for \( x \), \( y \), and \( z \)

Next, we replace the corresponding columns of \( \mathbf{A} \) with \( \mathbf{b} \) and compute the determinants to find \( x \), \( y \), and \( z \).

- **For \( x \):**
\[
\mathbf{A}_x = \begin{pmatrix} 5 & 1 & 1 \\ 3 & -2 & 2 \\ 4 & 2 & -1 \end{pmatrix}
\]
\[
\det(\mathbf{A}_x) = 5 \cdot \begin{vmatrix} -2 & 2 \\ 2 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 2 \\ 4 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & -2 \\ 4 & 2 \end{vmatrix}
\]
We already know \( \begin{vmatrix} -2 & 2 \\ 2 & -1 \end{vmatrix} = -2 \). Now calculate the other two:
\[
\det \begin{vmatrix} 3 & 2 \\ 4 & -1 \end{vmatrix} = (3)(-1) - (2)(4) = -3 - 8 = -11
\]
\[
\det \begin{vmatrix} 3 & -2 \\ 4 & 2 \end{vmatrix} = (3)(2) - (-2)(4) = 6 + 8 = 14
\]
Now substitute:
\[
\det(\mathbf{A}_x) = 5(-2) - 1(-11) + 1(14) = -10 + 11 + 14 = 15
\]

- **For \( y \):**
\[
\mathbf{A}_y = \begin{pmatrix} 1 & 5 & 1 \\ 3 & 3 & 2 \\ 2 & 4 & -1 \end{pmatrix}
\]
\[
\det(\mathbf{A}_y) = 1 \cdot \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} - 5 \cdot \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix}

Solve the following system of equations by Cramer’s rule: x + y + z = 5, 3x − 2y + 2z = 3, 2x + 2y − z = 4

This page explains how to solve the system of equations x + y + z = 5, 3x − 2y + 2z = 3, 2x + 2y − z = 4 using Cramer's Rule. Follow a detailed, step-by-step approach to finding the solution through matrix determinants, suitable for Grades 10-12.

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