Key Concept: Solving Systems of Linear Equations using Determinants (Cramer's Rule)

For a system of linear equations with $n$ variables, represented in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix, we can use determinants to determine the nature of its solutions.

Let $D = \det(A)$ be the determinant of the coefficient matrix. Let $D_j$ be the determinant obtained by replacing the $j$-th column of $A$ with the constant matrix $B$.

The following conditions apply:

1. Unique Solution: If $D \ne 0$, the system has a unique solution given by $x_j = \frac{D_j}{D}$ for all $j$.
2. No Solution (Inconsistent System): If $D = 0$ AND at least one of $D_1, D_2, D_3, \dots, D_n$ is non-zero, then the system has no solution.
3. Infinitely Many Solutions (Consistent System with Dependent Equations): If $D = 0$ AND all of $D_1, D_2, D_3, \dots, D_n$ are zero, then the system has infinitely many solutions.

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Step-by-Step Solution

1. Represent the System in Matrix Form The given system of linear equations is: $x_1 + 2x_2 + x_3 = 3$ $2x_1 + 3x_2 + x_3 = 3$ $3x_1 + 5x_2 + 2x_3 = 1$

We can write this in the matrix form $AX = B$, where: $A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \quad B = \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix}$

2. Calculate the Determinant of the Coefficient Matrix, $D$ First, we calculate the determinant of the coefficient matrix $A$, denoted as $D$. This is the crucial first step to determine if a unique solution exists. $D = \det(A) = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{vmatrix}$ We can evaluate this determinant using cofactor expansion along the first row: $D = 1 \cdot \begin{vmatrix} 3 & 1 \\ 5 & 2 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 3 \\ 3 & 5 \end{vmatrix}$ $D = 1 \cdot ((3)(2) - (1)(5)) - 2 \cdot ((2)(2) - (1)(3)) + 1 \cdot ((2)(5) - (3)(3))$ $D = 1 \cdot (6 - 5) - 2 \cdot (4 - 3) + 1 \cdot (10 - 9)$ $D = 1 \cdot (1) - 2 \cdot (1) + 1 \cdot (1)$ $D = 1 - 2 + 1$ $D = 0$ Explanation: Since $D = 0$, the system does not have a unique solution. It implies that either there are no solutions (inconsistent system) or infinitely many solutions (consistent system with dependent equations). To distinguish between these two cases, we need to calculate at least one of $D_1, D_2, D_3$.

3. Calculate the Determinant $D_1$ Next, we calculate $D_1$, which is the determinant of the matrix formed by replacing the first column of $A$ with the constant terms from matrix $B$. $D_1 = \begin{vmatrix} 3 & 2 & 1 \\ 3 & 3 & 1 \\ 1 & 5 & 2 \end{vmatrix}$ We evaluate this determinant using cofactor expansion along the first row: $D_1 = 3 \cdot \begin{vmatrix} 3 & 1 \\ 5 & 2 \end{vmatrix} - 2 \cdot \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & 3 \\ 1 & 5 \end{vmatrix}$ $D_1 = 3 \cdot ((3)(2) - (1)(5)) - 2 \cdot ((3)(2) - (1)(1)) + 1 \cdot ((3)(5) - (3)(1))$ $D_1 = 3 \cdot (6 - 5) - 2 \cdot (6 - 1) + 1 \cdot (15 - 3)$ $D_1 = 3 \cdot (1) - 2 \cdot (5) + 1 \cdot (12)$ $D_1 = 3 - 10 + 12$ $D_1 = 5$ Explanation: We found that $D_1 = 5$, which is non-zero.

4. Conclude the Nature of the Solution We have $D = 0$ and $D_1 \ne 0$. According to Cramer's Rule conditions, when $D=0$ and at least one of $D_j$ is non-zero, the system of equations is inconsistent, meaning it has no solution.

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Tips and Common Mistakes:

• Careful Calculation: Determinant calculations can be prone to arithmetic errors. Double-check your work, especially signs during cofactor expansion.
• Understanding Conditions: Memorize the conditions for unique, no, and infinite solutions based on the values of $D$ and $D_j$. A common mistake is to stop after finding $D=0$ and immediately conclude "no solution" without checking $D_j$.
• Alternative Methods: While Cramer's Rule is powerful for determining the nature of solutions, Gaussian elimination (row operations on the augmented matrix) is another robust method that can also find the solutions (if they exist) and clearly show inconsistencies or dependencies. For competitive exams, choose the method you are most proficient with and that seems quickest for the given problem.
• "Exactly 3 solutions" is not possible for a linear system: A system of linear equations can only have a unique solution, no solution, or infinitely many solutions. It can never have a finite number of solutions other than one.

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Summary and Key Takeaway We analyzed the given system of linear equations using the determinant method (Cramer's Rule).

1. The determinant of the coefficient matrix, $D$, was found to be $0$. This indicates that the system does not have a unique solution.
2. The determinant $D_1$ (formed by replacing the first column of coefficients with the constant terms) was found to be $5$, which is non-zero.
3. The condition $D = 0$ and at least one $D_j \ne 0$ implies that the system of linear equations is inconsistent and therefore has no solution.

The final answer is $\boxed{\text{no solution}}$.