Problem 17 - Algebra for College Students

Question

Use a matrix approach to solve each system. \(\left(\begin{array}{l}3 x-5 y=39 \\ 2 x+7 y=-67\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \(x = -9.5\) and \(y = -13.5\).

Step 1: Identify the Coefficients

First, identify the coefficients from the given system of equations. We have two equations: 1. \(3x - 5y = 39\)2. \(2x + 7y = -67\)The coefficients form the following matrix:\[ A = \begin{bmatrix} 3 & -5 \ 2 & 7 \end{bmatrix} \]

Step 2: Form the Matrices

Create the augmented matrix for this system, which includes the coefficients and the constants from each equation. This is written as:\[ \begin{bmatrix} 3 & -5 & | & 39 \ 2 & 7 & | & -67 \end{bmatrix} \]

Step 3: Reduce to Row Echelon Form

Use Gaussian elimination to convert the augmented matrix to row-echelon form.First, use row operations to make the element at the 1st row, 1st column to be 1. Divide the first row by 3:\[ \begin{bmatrix} 1 & -\frac{5}{3} & 13 \ 2 & 7 & -67 \end{bmatrix} \]Next, eliminate the 2 in row 2, column 1 by replacing row 2 with (row 2 - 2 * row 1):\[ \begin{bmatrix} 1 & -\frac{5}{3} & 13 \ 0 & \frac{26}{3} & -93 - 26 \end{bmatrix} \]This simplifies to:\[ \begin{bmatrix} 1 & -\frac{5}{3} & 13 \ 0 & \frac{26}{3} & -119 \end{bmatrix} \]

Step 4: Back Substitution to Find Solutions

Now solve the system from the row echelon form matrix:The second row represents the equation:\[ \frac{26}{3}y = -119 \]Solve for \(y\):\[ y = -\frac{119 \times 3}{26} = -13.5 \]Substitute \(y = -13.5\) back into the first equation:\[ x - \frac{5}{3}(-13.5) = 13 \]Simplify:\[ x + 22.5 = 13 \]Solving for \(x\), we have:\[ x = 13 - 22.5 = -9.5 \]

Step 5: Write the Solution

Write the solution of the system based on the results from the back substitution:\[ x = -9.5, \quad y = -13.5 \]

Key Concepts

Gaussian elimination

Gaussian elimination is a versatile method for solving systems of linear equations, widely used in mathematics due to its systematic approach. It involves transforming a system of simultaneous linear equations into an equivalent but simpler system that is easier to solve.

The process begins with the formation of an augmented matrix from the coefficients and constant terms of the equations. The aim of Gaussian elimination is to manipulate the matrix so that it achieves a triangular form, known as the row-echelon form. This is done through a series of row operations, which include swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting multiples of one row to another to eliminate elements below the pivot positions.

• Swapping rows can place larger numbers at the top, making later steps easier.
• Scaling still allows us to keep equivalent systems.
• Row addition or subtraction helps in eliminating variables.

Once the matrix is in row-echelon form, the solutions can be found through a process called back substitution.

augmented matrix

An augmented matrix is a fundamental part of the Gaussian elimination process. It is a compact representation that consolidates both the coefficients of variables and the constants from the equations into a single matrix. This allows for streamlined operations when solving systems.

The augmented matrix includes everything you need to know about a system of equations in one handy layout. Take, for example, the system of equations:

• Equation 1: \(3x - 5y = 39\)
• Equation 2: \(2x + 7y = -67\)

The coefficients form the matrix: \[\begin{bmatrix}3 & -5 \2 & 7\end{bmatrix}\] And the augmented matrix, which also includes the constant terms, is:\[\begin{bmatrix}3 & -5 & \vert & 39 \2 & 7 & \vert & -67 \end{bmatrix}\]This format allows for easier application of row operations to achieve the desired row-echelon form.

row-echelon form

Row-echelon form is a target stage in the process of Gaussian elimination where the matrix is simplified to make back substitution straightforward. A matrix is in row-echelon form when:

• All non-zero rows are above rows of all zeros.
• The leading entry of each non-zero row (known as the pivot) is 1 and is to the right of the leading entry in the row above.
• Each column containing a pivot has zeros in all its other entries.

For example, consider the transformation of our augmented matrix to row-echelon form:\[\begin{bmatrix}1 & -\frac{5}{3} & 13 \0 & \frac{26}{3} & -119\end{bmatrix}\]Here, the matrix is simplified such that the pivot (1 in the first row's first column) is above its respective zero position (0 in the second row). To reach this form, we performed operations like row swapping and scaling, allowing easier determination of variable values.

back substitution

Back substitution is the final step after achieving the row-echelon form during Gaussian elimination. It involves back-tracing through the matrix to find the values of the variables.

Given the row-echelon form: \[\begin{bmatrix}1 & -\frac{5}{3} & 13 \0 & \frac{26}{3} & -119\end{bmatrix}\]We interpret these as simplified equations. The second row directly gives us one of the variables:\(\frac{26}{3}y = -119\) Solving for \(y\) involves multiplication and division to isolate \(y\). Once \(y\) is found, we can substitute it back into the first row's equation, which provides the value for \(x\). This method ensures that all variable values satisfy the original system of equations, giving us a complete solution. The solutions in our case are \(x = -9.5\) and \(y = -13.5\), derived systematically through this process.