Problem 58 - 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. $$ \left\\{\begin{array}{r} x+2 y=0 \\ x+y=6 \\ 3 x-2 y=8 \end{array}\right. $$

Step-by-Step Solution

Answer

The system of equations has no solutions.

Step 1: Augmented Matrix Formation

First, translate this system of equations into an augmented matrix. The coefficients of \(x\), \(y\) and the constants on the other side of equality form this matrix. So, for the given system of equations, the augmented matrix would be: \[ \left[ \begin{array}{cc|c} 1 & 2 & 0 \ 1 & 1 & 6 \ 3 & -2 & 8 \end{array} \right] \]

Step 2: Gaussian Elimination

Now, perform the Gaussian elimination. We aim to create zeroes in the first column below the first row. Subtract the first row from the second, and subtract three times the first row from the third: \[ \left[ \begin{array}{cc|c} 1 & 2 & 0 \ 0 & -1 & 6 \ 0 & -8 & 8 \end{array} \right] \] Next, swap the 2nd and 3rd row, and then multiply the new second row by -1 to simplify: \[ \left[ \begin{array}{cc|c} 1 & 2 & 0 \ 0 & 8 & 8 \ 0 & -1 & 6 \end{array} \right] \rightarrow \left[ \begin{array}{cc|c} 1 & 2 & 0 \ 0 & 1 & 1 \ 0 & -1 & 6 \end{array} \right] \] Finally add the new second row to the third row: \[ \left[ \begin{array}{cc|c} 1 & 2 & 0 \ 0 & 1 & 1 \ 0 & 0 & 7 \end{array} \right] \]

Step 3: Back-substitution

Use back-substitution to find the values of \(x\) and \(y\). The last row equates to '0z = 7', which has no solutions, implying that the given system of eqations has no solutions.

Key Concepts

System of Equations

A system of equations consists of two or more equations that share the same set of variables. In this exercise, the system includes three equations with two variables, \( x \) and \( y \):

• \( x + 2y = 0 \)
• \( x + y = 6 \)
• \( 3x - 2y = 8 \)

These equations represent mathematical relationships that must all be true simultaneously. The goal is to find a set of values for \( x \) and \( y \) that satisfy all equations at the same time. Solving a system can lead to a unique solution, infinitely many solutions, or no solution at all. Here, we find that the system has no solution. This means there is no point \( (x, y) \) that satisfies all three equations together.

Augmented Matrix

To solve a system of equations using matrices, it's helpful to convert the equations into an augmented matrix. The augmented matrix encapsulates the coefficients and constant terms of the equations. For our system:

• First row: coefficients from \( x + 2y = 0 \) become \([1, 2 | 0]\)
• Second row: coefficients from \( x + y = 6 \) become \([1, 1 | 6]\)
• Third row: coefficients from \( 3x - 2y = 8 \) become \([3, -2 | 8]\)

The vertical line in the matrix \( \left[ \begin{array}{cc|c} 1 & 2 & 0 \ 1 & 1 & 6 \ 3 & -2 & 8 \end{array} \right] \) separates the coefficients of the variables from the constant terms. This matrix form is crucial for applying methods like Gaussian elimination, as it provides a structured way to manipulate multiple equations simultaneously.

Matrices in Algebra

Matrices play a central role in algebra for solving systems of equations, handling transformations, and representing data. In the context of solving linear systems, matrices are employed to systematically and efficiently manage the computations required to find solutions.

• Gaussian Elimination: This involves a series of operations (row swaps, scaling, and row additions or subtractions) on the augmented matrix to drive it to a form where solutions can be easily read off. It aims to form an upper triangular matrix.
• Back-Substitution: Once the matrix is in upper triangular form, back-substitution allows for extracting the values of unknowns starting from the bottom row up.

These operations ensure consistency across multiple equations, offering a clear path to determine whether there are solutions and what they might be. In our exercise, the result indicated 'no solution', which is one of the possible outcomes when solving systems with matrices.