Question

By forming the augmented matrix and row reducing, determine the solutions of the following system $$ \begin{aligned} &2 x-y+3 z=4 \\ &3 x+2 z=5 \\ &-2 x+y+4 z=6 \end{aligned} $$

Short answer

The solutions to the given system of equations are \(x = -1\), \(y = -\frac{26}{3}\), and \(z = 5\).

Step-by-step solution

Write down the augmented matrix

We form the augmented matrix by taking the coefficients of the variables and the constants in the system: \( \left[ \begin{array}{ccc|c} 2 & -1 & 3 & 4 \\ 3 & 0 & 2 & 5 \\ -2 & 1 & 4 & 6 \end{array} \right] \)

Row reduce the matrix

We will perform row reduction (Gaussian elimination) to get an equivalent matrix in row-echelon form, then continue to reduced row-echelon form (RREF). 1. First, swap the first and second rows to get a matrix with a leading 1: \( \left[ \begin{array}{ccc|c} 3 & 0 & 2 & 5 \\ 2 & -1 & 3 & 4 \\ -2 & 1 & 4 & 6 \end{array} \right] \) 2. Next, divide the first row by 3 to get a leading 1: \( \left[ \begin{array}{ccc|c} 1 & 0 & \frac{2}{3} & \frac{5}{3} \\ 2 & -1 & 3 & 4 \\ -2 & 1 & 4 & 6 \end{array} \right] \) 3. Now, eliminate the first column by subtracting 2 times the first row from the second row, and adding 2 times the first row to the third row: \( \left[ \begin{array}{ccc|c} 1 & 0 & \frac{2}{3} & \frac{5}{3} \\ 0 & -1 & \frac{7}{3} & \frac{7}{3} \\ 0 & 1 & \frac{14}{3} & \frac{22}{3} \end{array} \right] \) 4. To proceed, change the second row to have a leading 1 by multiplying it by -1: \( \left[ \begin{array}{ccc|c} 1 & 0 & \frac{2}{3} & \frac{5}{3} \\ 0 & 1 & -\frac{7}{3} & -\frac{7}{3} \\ 0 & 1 & \frac{14}{3} & \frac{22}{3} \end{array} \right] \) 5. Eliminate the third column by adding the second row to the third row: \( \left[ \begin{array}{ccc|c} 1 & 0 & \frac{2}{3} & \frac{5}{3} \\ 0 & 1 & -\frac{7}{3} & -\frac{7}{3} \\ 0 & 0 & \frac{7}{3} & \frac{15}{3} \end{array} \right] \) 6. Finally, change the third row to have a leading 1 by multiplying it by \(\frac{3}{7}\): \( \left[ \begin{array}{ccc|c} 1 & 0 & \frac{2}{3} & \frac{5}{3} \\ 0 & 1 & -\frac{7}{3} & -\frac{7}{3} \\ 0 & 0 & 1 & \frac{5}{1} \end{array} \right] \)

Solve for each variable

Now that the matrix is in reduced row-echelon form, we can read the solutions for the variables directly: $$ z = \frac{5}{1} = 5 $$ $$ y = -\frac{7}{3}z - \frac{7}{3} = -\frac{26}{3} $$ $$ x = \frac{5}{3} - \frac{2}{3}z = \frac{5}{3} - \frac{2}{3}(5) = -1 $$ So, the solutions to the system of equations are \(x = -1\), \(y = -\frac{26}{3}\), and \(z = 5\).

Key concepts

Augmented Matrix

In any system of linear equations, each equation represents a linear relationship among variables. To solve these efficiently, we convert the system into a matrix format called the augmented matrix. This matrix consists of two main parts: the matrix of coefficients and the constant matrix. Each row in the augmented matrix correlates to a single equation from the original system.

For our exercise, the equations are transformed into the following augmented matrix:

• The first row: The coefficients are 2, -1, and 3 for variables x, y, and z respectively, followed by the constant 4 from the equation, forming:
• The second row uses the coefficients 3, 0, and 2 for x, y, and z, with constant 5:
• The third row has -2, 1, and 4, followed by constant 6:

This arrangement allows us to use matrix operations, simplifying and solving the system more effectively.

Row Reduction Techniques

Row Reduction, or Gaussian Elimination, transforms an augmented matrix into a simpler form to solve the corresponding system of equations. There are three basic "row operations" employed to manipulate the matrix without altering the solution of the system:

• Swapping two rows.
• Multiplying a row by a scalar (non-zero).
• Adding or subtracting a multiple of one row to/from another row.

In our example, these techniques help achieve a structured form where zeros are systematically introduced below and sometimes above leading coefficients (generally 1's).
After swapping and scaling rows, we aim to create zeros in the columns below the leading entries, systematically simplifying our systems. This structured format, known as echelon form, greatly facilitates the subsequent computations needed to reach reduced row-echelon form.

Reduced Row-Echelon Form

Reduced Row-Echelon Form (RREF) is a refined stage of matrix simplification. It represents a state where the augmented matrix outlines the solutions to the system of equations directly. To achieve RREF, the matrix must satisfy certain conditions:

• Leading entries (first non-zero number from the left) in each row must be 1.
• Each leading 1 is the only non-zero entry in its column.
• Each successive leading 1 occurs in a column further to the right than the one before it.
• Any row containing only zeros is at the bottom.

Our final transformation of the matrix achieves this ideal form, allowing direct interpretation of solutions for each variable. For instance:

• The third row shows directly that z = 5.
• The second row, as expanded, relates to the solution of y.
• Finally, the first row provides the value of x after substituting in known values of z and y.

This straightforward form simplifies the once complex system into a set of straightforward calculations.

Solving Systems of Equations

Solving a system of equations with matrices reduces the complexity of simultaneously considering multiple variables and equations. By converting the system into an enhanced matrix form (reduced row-echelon form), we can directly solve for each variable starting from the last row upwards, a method known as back substitution.

In this exercise:

• We first determine the value of z from the last row which straightforwardly gives us z = 5.
• We use the value of z in the second row equation to find y:
• The equation simplifies to evaluate y easily.
• Finally, substituting known values of y and z in the first row yields x.

This back substitution technique simplifies complex calculations, allowing a logical progression through the system, ensuring accurate and clear solutions from beginning to end.