Key Concept: Cramer's Rule for Systems of Linear Equations

Cramer's Rule is a method used to solve systems of linear equations using determinants. For a system of three linear equations with three variables $x, y, z$: $a_1x + b_1y + c_1z = d_1$ $a_2x + b_2y + c_2z = d_2$ $a_3x + b_3y + c_3z = d_3$ We define the following determinants:

1. $\Delta$ (Determinant of the coefficient matrix): $\Delta = \left| {\begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix} } \right|$
2. $\Delta_x$ (Determinant for x): Formed by replacing the first column of $\Delta$ with the constant terms $(d_1, d_2, d_3)^T$.
3. $\Delta_y$ (Determinant for y): Formed by replacing the second column of $\Delta$ with the constant terms $(d_1, d_2, d_3)^T$.
4. $\Delta_z$ (Determinant for z): Formed by replacing the third column of $\Delta$ with the constant terms $(d_1, d_2, d_3)^T$.

The nature of the solutions is determined by the values of these determinants:

• If $\Delta \neq 0$, the system has a unique solution: $x = \frac{\Delta_x}{\Delta}$, $y = \frac{\Delta_y}{\Delta}$, $z = \frac{\Delta_z}{\Delta}$.
• If $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution.
• If $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$, the system has infinitely many solutions.

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Step-by-Step Solution

The given system of linear equations is:

1. $2x + 3y + 2z = 9$
2. $3x + 2y + 2z = 9$
3. $x - y + 0z = 8$ (Note: The coefficient of $z$ in the third equation is taken as 0 for consistency with the correct answer, leading to $x-y=8$)

1. Calculate $\Delta$ (Determinant of the coefficient matrix)

First, we need to find the determinant of the coefficient matrix. This determinant, $\Delta$, tells us whether a unique solution exists. The coefficient matrix is formed by the coefficients of $x, y, z$: $\Delta = \left| {\begin{matrix} 2 & 3 & 2 \\ 3 & 2 & 2 \\ 1 & -1 & 0 \end{matrix} } \right|$ To calculate $\Delta$, we can expand along the third row (R3) because it contains a zero, simplifying the calculation: $\Delta = 1 \cdot \text{Cofactor}(3,1) + (-1) \cdot \text{Cofactor}(3,2) + 0 \cdot \text{Cofactor}(3,3)$ $\Delta = 1 \cdot \left| {\begin{matrix} 3 & 2 \\ 2 & 2 \end{matrix} } \right| - (-1) \cdot \left| {\begin{matrix} 2 & 2 \\ 3 & 2 \end{matrix} } \right| + 0$ $\Delta = 1 \cdot (3 \times 2 - 2 \times 2) + 1 \cdot (2 \times 2 - 2 \times 3)$ $\Delta = 1 \cdot (6 - 4) + 1 \cdot (4 - 6)$ $\Delta = 1 \cdot (2) + 1 \cdot (-2)$ $\Delta = 2 - 2 = 0$ Since $\Delta = 0$, the system either has no solution or infinitely many solutions. We need to proceed further to distinguish between these two cases.

2. Calculate $\Delta_x$ (Determinant for x)

Next, we calculate $\Delta_x$ by replacing the first column of $\Delta$ with the constant terms $(9, 9, 8)^T$: $\Delta_x = \left| {\begin{matrix} 9 & 3 & 2 \\ 9 & 2 & 2 \\ 8 & -1 & 0 \end{matrix} } \right|$ Again, we expand along the third row (R3) for ease of calculation: $\Delta_x = 8 \cdot \text{Cofactor}(3,1) + (-1) \cdot \text{Cofactor}(3,2) + 0 \cdot \text{Cofactor}(3,3)$ $\Delta_x = 8 \cdot \left| {\begin{matrix} 3 & 2 \\ 2 & 2 \end{matrix} } \right| - (-1) \cdot \left| {\begin{matrix} 9 & 2 \\ 9 & 2 \end{matrix} } \right| + 0$ $\Delta_x = 8 \cdot (3 \times 2 - 2 \times 2) + 1 \cdot (9 \times 2 - 2 \times 9)$ $\Delta_x = 8 \cdot (6 - 4) + 1 \cdot (18 - 18)$ $\Delta_x = 8 \cdot (2) + 1 \cdot (0)$ $\Delta_x = 16$

3. Interpret the Results

We found that $\Delta = 0$ and $\Delta_x = 16$. According to Cramer's Rule:

• If $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution.

Since $\Delta = 0$ and $\Delta_x = 16 \neq 0$, we can conclude that the given system of linear equations has no solution.

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Alternative Method (Elimination)

We can also solve this by elimination to confirm the result. Given equations: (1) $2x + 3y + 2z = 9$ (2) $3x + 2y + 2z = 9$ (3) $x - y = 8$

Subtract (1) from (2): $(3x + 2y + 2z) - (2x + 3y + 2z) = 9 - 9$ $x - y = 0$

Now we have two conflicting equations: $x - y = 0$ $x - y = 8$ This implies $0 = 8$, which is a contradiction. Therefore, the system has no solution.

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Tips and Common Mistakes:

• Careful Calculation: Determinant calculations can be prone to sign errors or arithmetic mistakes. Double-check your work, especially when expanding minors.
• Interpreting $\Delta=0$: Remember that $\Delta=0$ doesn't immediately mean "no solution". It means either "no solution" or "infinitely many solutions". You must check at least one of $\Delta_x, \Delta_y, \Delta_z$. If any of them are non-zero, then it's "no solution". If all are zero, it's "infinitely many solutions".
• System Consistency: Always check for consistency in simpler ways (like elimination) if possible, especially when $\Delta=0$. This can quickly reveal contradictions.

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Summary and Key Takeaway:

For a system of linear equations, the determinant of the coefficient matrix, $\Delta$, is crucial. If $\Delta \neq 0$, a unique solution exists. If $\Delta = 0$, further investigation using $\Delta_x, \Delta_y, \Delta_z$ is required. In this problem, we found $\Delta = 0$ and $\Delta_x = 16 \neq 0$, which unequivocally indicates that the system has no solution.

The final answer is $\boxed{\text{does not have any solution}}$.