Problem 60 - College Algebra and Calculus: An Applied Approach

Question

Use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} x-3 y &=5 \\ -2 x+6 y &=-10 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 5 + 3y\), where \(y\) is any real number.

Step 1: Write the System of Equations in Matrix Form

The system of equations can be expressed in the matrix form \(Ax = b\), where \(A\) is the matrix of coefficients, \(x\) is the vector of variables, and \(b\) is the vector of constants. For this particular problem, the matrix, \(A\), is \[\begin{bmatrix}1 & -3\\ -2 & 6\end{bmatrix}\] and \(b\) is \[\begin{bmatrix}5\\ -10\end{bmatrix}\].

Step 2: Apply Gaussian Elimination

The goal of Gaussian elimination is to transform the matrix into row-echelon form, where the lower-left entries are all zero. Since we already have a 1 in the top left position, we can scale the first equation by 2 and add it to the second equation to make the equation below zero, which would give us the row-echelon form. This will result in the following augmented matrix: \[\begin{bmatrix}1 & -3 & | & 5\\ 0 & 0 & | & 0\end{bmatrix}\]

Step 3: Understand the Resulting Matrix is a Sign of Infinite Solutions

The resulting matrix from the Gaussian elimination shows that the second row consists of 0's. This means that the two given equations are dependent and correspond to the same line, suggesting an infinite number of solutions. The solution to this system of equations is \(x = 5 + 3y\), where \(y\) is any real number.

Key Concepts

Gaussian Elimination

Gaussian elimination is a method used in linear algebra to solve systems of linear equations. It involves a series of operations on the augmented matrix to transform it into a simpler form. The process is named after the German mathematician Carl Friedrich Gauss.
The main goal is to convert the matrix into row-echelon form. This form has the leading entry in each row positioned to the right of the leading entry in the row above it, and the entries below these leading entries are zeros.
Key steps in Gaussian elimination:

• Swap rows to get a non-zero entry in the top left corner, if needed.
• Scale rows to make the pivot (the leading entry) equal to 1.
• Use the pivot to eliminate all entries below it in the current column.
• Repeat these steps for each row moving from top to bottom and column to column.

In our example, the pivot entry was already 1, allowing us to quickly apply the method to reach the row-echelon form. This setup makes it easier to determine the properties of the solution set, helping us distinguish between unique, infinite, or no solutions.

System of Equations

A system of equations consists of multiple equations that are solved together because they share common variables. It can be represented in matrix form, which helps in applying techniques such as Gaussian elimination.
In the given exercise, the system is displayed as:\[\begin{aligned} x-3y &= 5 \-2x+6y &= -10 \end{aligned}\]This system can be expressed in a simplified linear matrix equation form, \(Ax = b\), where

• \(A\) is the coefficient matrix \(\begin{bmatrix}1 & -3 \ -2 & 6\end{bmatrix}\)
• \(x\) is the column vector of variables: \(\begin{bmatrix}x \ y\end{bmatrix}\)
• \(b\) is the constant matrix \(\begin{bmatrix}5 \ -10\end{bmatrix}\)

The steps of transformation and operations should preserve the equivalence of the original system. When manipulated correctly, the matrix helps to directly observe the relationships between the equations.

Infinite Solutions

Infinite solutions occur in a system of linear equations when the equations represent the same line in graphical space, meaning they have all their solutions in common.
After applying Gaussian elimination in our problem, the augmented matrix transformed into:\[\begin{bmatrix}1 & -3 & | & 5\ 0 & 0 & | & 0\end{bmatrix}\]The second row of zeroes indicates that the two original equations are dependent. This means any solution to one equation is also a solution to the other.
The general solution can then be expressed in terms of one of the variables. Here, \(x\) is expressed in terms of \(y\):
\(x = 5 + 3y\), making \(y\) any real number, thus producing infinite solutions. This concept highlights the dependency of equations and how matrix methods can quickly reveal such properties.