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When solving a system of linear equations, one efficient method is to use the matrix form. This involves expressing the system in terms of matrices, which can simplify and organize the calculations. For example, consider the system of equations:\[ \begin{aligned} 4x - y + 2z &= 1 \ 3x + 5y - z &= -2 \ -9x - 15y + 3z &= 0 \end{aligned} \]To express this system in matrix form, place the coefficients of the variables into a matrix, known as the coefficient matrix, and set up the augmented matrix with the constants on the right-hand side. The matrix form looks like this:\[ \begin{bmatrix} 4 & -1 & 2 \ 3 & 5 & -1 \ -9 & -15 & 3 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ -2 \ 0 \end{bmatrix} \] This matrix form allows for a systematic approach to solving the system using techniques like row reduction.

Row reduction, also known as Gaussian elimination, is a method for solving systems of linear equations. It involves performing operations on the rows of the augmented matrix to achieve a simpler form. Let's break down the row reduction steps used to solve our example.First, write the augmented matrix:\[ \begin{bmatrix} 4 & -1 & 2 & | & 1 \ 3 & 5 & -1 & | & -2 \ -9 & -15 & 3 & | & 0 \end{bmatrix} \]Performing row operations:- Make the leading coefficient (top left corner) a 1 by dividing Row 1 by 4:\[ \begin{bmatrix} 1 & -\frac{1}{4} & \frac{1}{2} & | & \frac{1}{4} \ 3 & 5 & -1 & | & -2 \ -9 & -15 & 3 & | & 0 \end{bmatrix} \]- Eliminate the first column below the leading 1 by subtracting 3 times Row 1 from Row 2 and adding 9 times Row 1 to Row 3:\[ \begin{bmatrix} 1 & -\frac{1}{4} & \frac{1}{2} & | & \frac{1}{4} \ 0 & 5.75 & -\frac{5}{2} & | & -2.75 \ 0 & -12.75 & 7.5 & | & 2.25 \end{bmatrix} \]- Make the leading coefficient in Row 2 a 1 by dividing Row 2 by 5.75:\[ \begin{bmatrix} 1 & -\frac{1}{4} & \frac{1}{2} & | & \frac{1}{4} \ 0 & 1 & -\frac{10}{23} & | & -\frac{11}{23} \ 0 & -12.75 & 7.5 & | & 2.25 \end{bmatrix} \]- Eliminate the second column below the leading 1 by adding 12.75 times Row 2 to Row 3:\[ \begin{bmatrix} 1 & -\frac{1}{4} & \frac{1}{2} & | & \frac{1}{4} \ 0 & 1 & -\frac{10}{23} & | & -\frac{11}{23} \ 0 & 0 & 0 & | & -\frac{75}{23} \end{bmatrix} \]Through row reduction, the system is simplified, making it easier to analyze the number of solutions.

To determine the number of solutions of a system of equations, we examine the final form of the augmented matrix after row reduction. The possible outcomes are:- **Unique Solution:** If the matrix reduces to a form where each variable can be solved uniquely.- **Infinite Solutions:** If the matrix reduces such that there are free variables, leading to multiple solutions.- **No Solution:** If a row reduces to a contradiction (e.g., 0 = non-zero number).In our example, the final matrix after row reduction is:\[ \begin{bmatrix} 1 & -\frac{1}{4} & \frac{1}{2} & | & \frac{1}{4} \ 0 & 1 & -\frac{10}{23} & | & -\frac{11}{23} \ 0 & 0 & 0 & | & -\frac{75}{23} \end{bmatrix} \]The third row translates to the equation 0 = -\frac{75}{23}, which is a contradiction. This tells us there is no solution for this system. When analyzing a system of equations, always check for contradictions or free variables to determine the number of possible solutions.