Question

Solve the following sets of equations by reducing the matrix to row echelon form. \(\left\\{\begin{array}{l}2 x+y=4 \\ 7 x-2 y=3\end{array}\right.\)

Short answer

x = \frac{2}{9}, y = \frac{22}{9}

Step-by-step solution

- Write the system as an augmented matrix

First, write the system of linear equations as an augmented matrix. Represent the system \[ \begin{cases} 2x + y = 4 \ 7x - 2y = 3 \end{cases} \]as the augmented matrix\[ \begin{pmatrix} 2 & 1 & | & 4 \ 7 & -2 & | & 3 \ \end{pmatrix} \]

- Perform row operations to get a leading 1 in the first column

Divide the first row (R1) by 2 to get the leading 1: \[ R1 \rightarrow \frac{1}{2} R1 \]The updated matrix is:\[ \begin{pmatrix} 1 & \frac{1}{2} & | & 2 \ 7 & -2 & | & 3 \ \end{pmatrix} \]

- Eliminate the first column's entry in the second row

Subtract 7 times the first row (R1) from the second row (R2): \[ R2 \rightarrow R2 - 7R1 \]This gives:\[ \begin{pmatrix} 1 & \frac{1}{2} & | & 2 \ 0 & -\frac{9}{2} & | & -11 \ \end{pmatrix} \]

- Scale the second row to get a leading 1

Divide the second row (R2) by -\(\frac{9}{2}\):\[ R2 \rightarrow \frac{2}{-9} R2 \]The matrix becomes:\[ \begin{pmatrix} 1 & \frac{1}{2} & | & 2 \ 0 & 1 & | & \frac{22}{9} \ \end{pmatrix} \]

- Eliminate the second column's entry in the first row

Subtract \(\frac{1}{2}\) times the second row (R2) from the first row (R1):\[ R1 \rightarrow R1 - \frac{1}{2}R2 \]This results in:\[ \begin{pmatrix} 1 & 0 & | & \frac{2}{9} \ \ 0 & 1 & | & \frac{22}{9} \ \end{pmatrix} \]

- Interpret the final matrix

The final matrix represents the equations:\[ \begin{cases} x = \frac{2}{9} \ y = \frac{22}{9} \end{cases} \]

Key concepts

Augmented Matrix

An augmented matrix is a way to represent a system of linear equations. It includes the coefficients of the variables and the constants from the equations. For example, for the system of equations \( 2x + y = 4 \) and \( 7x - 2y = 3 \), the augmented matrix is written as:\[ \begin{pmatrix} 2 & 1 & | & 4 \ 7 & -2 & | & 3 \ \end{pmatrix} \] The vertical line in the matrix separates the coefficients of the variables from the constants. This form makes it easier to apply matrix operations and find solutions.

Row Operations

Row operations are techniques used to manipulate the rows of a matrix to solve systems of linear equations. There are three main types of row operations:

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

These operations help us transform the matrix into Row Echelon Form (REF). For example, in step 2 of the solution, the first row is divided by 2, so that the first entry becomes 1:
\(R1 \rightarrow \frac{1}{2} R1 \)
The new matrix is:

\[ \begin{pmatrix} 1 & \frac{1}{2} & | & 2 \ 7 & -2 & | & 3 \end{pmatrix} \]

Linear Equations

Linear equations are equations where each term is either a constant or the product of a constant and a single variable. Linear systems consist of multiple linear equations. For example, take the equations:
\[ 2x + y = 4 \ \ 7x - 2y = 3 \ \]
These are linear because they are composed of constants and variables raised to the power of 1. To solve a system of linear equations, we can use matrix methods like the augmented matrix and row operations to simplify and find the values of the variables. The steps in the given exercise use these techniques to reduce the matrix to row echelon form and then interpret it to find the solutions \( x = \frac{2}{9} \) and \( y = \frac{22}{9} \).