Problem 71 - Precalculus

Question

Solve the system of linear equations using Gauss-Jordan elimination. $$\begin{array}{rr} x+3 y= & -5 \\ -2 x-y= & 0 \end{array}$$

Step-by-Step Solution

Answer

The solution is \( x = 1 \) and \( y = -2 \).

Step 1: Write the Augmented Matrix

The augmented matrix represents the system of equations in a matrix form. For the given system:\[ \begin{bmatrix} 1 & 3 & | & -5 \ -2 & -1 & | & 0 \end{bmatrix} \] represents the equations \( x + 3y = -5 \) and \( -2x - y = 0 \).

Step 2: Apply Row Operations to Get a Leading 1 in the First Row, First Column

The first row already has a leading 1 in the first column, so no changes are needed yet. The augmented matrix remains the same.\[ \begin{bmatrix} 1 & 3 & | & -5 \ -2 & -1 & | & 0 \end{bmatrix} \]

Step 3: Eliminate the First Variable in the Second Row

Use row operations to transform the first element of the second row to zero. Do this by adding 2 times the first row to the second row.\[ R_2 = R_2 + 2 \times R_1 \] transforms the matrix to:\[ \begin{bmatrix} 1 & 3 & | & -5 \ 0 & 5 & | & -10 \end{bmatrix} \]

Step 4: Get a 1 as the Leading Entry in the Second Row, Second Column

Divide the entire second row by 5 to make the leading coefficient 1.\[ R_2 = \frac{1}{5} R_2 \] transforms the matrix to:\[ \begin{bmatrix} 1 & 3 & | & -5 \ 0 & 1 & | & -2 \end{bmatrix} \]

Step 5: Eliminate the Second Variable in the First Row

Subtract 3 times the second row from the first row to eliminate the second variable in the first row.\[ R_1 = R_1 - 3 \times R_2 \] transforms the matrix to:\[ \begin{bmatrix} 1 & 0 & | & 1 \ 0 & 1 & | & -2 \end{bmatrix} \]

Step 6: Verify and Write the Solution

The matrix is now in reduced row-echelon form. Thus, the solution to the system of equations is visible from the matrix. It tells us that \( x = 1 \) and \( y = -2 \).

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations with multiple variables. The goal is to find the values of these variables that satisfy all the equations simultaneously. For example, in the given system: - \(x + 3y = -5\) - \(-2x - y = 0\) we are looking for the values of \(x\) and \(y\) that make both equations true at the same time. Solving a system like this can help in various real-world problems, like optimizing resources or finding intersection points of lines.

Augmented Matrix

To solve a system of linear equations using matrix methods, we first translate the equations into an augmented matrix. This matrix includes all the coefficients of the variables and the constants on the other side of the equal sign. For our system: \[\begin{bmatrix} 1 & 3 & | & -5 \ -2 & -1 & | & 0 \end{bmatrix}\] represents the equations. Here, each row corresponds to an equation, and each column represents coefficients of the variables. The vertical line separates the coefficients from the constants, helping us visualize the system more clearly.

Row Operations

Row operations are key transformations used to simplify matrices in methods like Gauss-Jordan elimination. These operations include:

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

Such operations help in transforming the matrix closer to the reduced row-echelon form. In our exercise, we used them to systematically eliminate variables by: - Making leading ones - Creating zeros below or above these leading ones.

Reduced Row-Echelon Form

Reduced row-echelon form (RREF) of a matrix is achieved when:

• Every leading entry of a row is 1
• Leading 1s are the only non-zero entries in their column
• Leading 1 of a row is to the right of the leading 1 of the row above
• All zero rows are at the bottom of the matrix

In RREF, the matrix gives immediate solutions to the system of equations. Our exercise culminated in the matrix: \[\begin{bmatrix} 1 & 0 & | & 1 \ 0 & 1 & | & -2 \end{bmatrix}\] which clearly shows \(x = 1\) and \(y = -2\), solving the system effectively.