Question

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{array}{r} 2 x+3 y=6 \\ x-y=\frac{1}{2} \end{array}\right. $$

Short answer

The solution is \( x = \frac{3}{2} \) and \( y = 1 \).

Step-by-step solution

Write the system in matrix form

First, express the given system of equations in terms of an augmented matrix:\[\begin{pmatrix}2 & 3 & | & 6 \1 & -1 & | & \frac{1}{2}\end{pmatrix}\]

Create a leading 1 in the first row

We already have a leading 1 in the second row first column, so we swap the first row with the second row:\[\begin{pmatrix}1 & -1 & | & \frac{1}{2} \2 & 3 & | & 6\end{pmatrix}\]

Eliminate the first column of the second row

Use row operations to make the element below the leading 1 in the first column, i.e., the first element of the second row, zero:Row 2 - 2 * Row 1:\[\begin{pmatrix}1 & -1 & | & \frac{1}{2} \0 & 5 & | & 5\end{pmatrix}\]

Create a leading 1 in the second row

Divide the entire second row by 5 to create a leading 1:\[\begin{pmatrix}1 & -1 & | & \frac{1}{2} \0 & 1 & | & 1\end{pmatrix}\]

Eliminate the second column of the first row

Use row operations to make the element above the leading 1 in the second row (i.e., the second element of the first row) zero:Row 1 + Row 2:\[\begin{pmatrix}1 & 0 & | & \frac{3}{2} \0 & 1 & | & 1\end{pmatrix}\]

Interpret the result

The resulting matrix corresponds to the system:\[\begin{array}{c}x = \frac{3}{2} \y = 1\end{array}\]Thus, the solution to the system of equations is \( x = \frac{3}{2} \) and \( y = 1 \).

Key concepts

Augmented Matrix

To solve a system of equations using matrices, we first need to convert the system into an augmented matrix. An augmented matrix combines the coefficients of the variables and the constants from the equations into a single matrix. For instance, the system of equations given is:
\[ 2x + 3y = 6 \ x - y = \frac{1}{2} \]
Its augmented matrix form looks like this: \[ \begin{pmatrix} 2 & 3 & | & 6 \ 1 & -1 & | & \frac{1}{2} \ \end{pmatrix} \] The vertical bar separates the coefficients from the constants. This matrix format makes row operations easier to apply.

Row Operations

Row operations help manipulate the augmented matrix to find the solution of the system. They include:

• Swapping rows
• Multiplying a row by a non-zero constant
• Adding or subtracting rows

Our goal is to make the matrix simpler while preserving the solutions. In Step 2, we swapped rows to get a leading 1 in the first column of the first row:
\[ \begin{pmatrix} 1 & -1 & | & \frac{1}{2} \ 2 & 3 & | & 6 \ \end{pmatrix} \] These operations are powerful tools for solving systems.

Leading 1

The 'leading 1' simplifies the matrix one step at a time. By turning leading coefficients of each row into 1, we create a reference point for eliminating other coefficients in the column. For example, in Step 4, we divided the second row by 5 to create a leading 1:
\[ \begin{pmatrix} 1 & -1 & | & \frac{1}{2} \ 0 & 1 & | & 1 \ \end{pmatrix} \] The first row already has a leading 1. This sets us up for further simplifying the matrix to find solutions.

Inconsistent System

An inconsistent system has no solutions. It happens when row operations lead to a contradiction, like 0 = 5. In this example, after performing our operations, we didn't encounter such a row, indicating there's no contradiction. Hence, the system is **consistent**. Recognizing inconsistencies early can save you time in solving.

Solution of System of Equations

Once the augmented matrix is in reduced form, we find the solutions by interpreting the matrix back into equations. From: \[ \begin{pmatrix} 1 & 0 & | & \frac{3}{2} \ 0 & 1 & | & 1 \ \end{pmatrix} \] This converts to:
\[ x = \frac{3}{2},\text{ } y = 1 \] Thus, the solution to our original equations is \ (x = \frac{3}{2},\text{ } y = 1) \. This process illustrates how matrix operations simplify solving complex systems.