Problem 77 - 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. $$ \left\\{\begin{array}{r} x+y+z=0 \\ 2 x+3 y+z=0 \\ 3 x+5 y+z=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The system of equations has infinitely many solutions, with \(x = 0\), \(y = z = s\), where \(s\) is any real number.

Step 1: Set Up the Augmented Matrix

First, set up the augmented matrix which corresponds to the system of linear equations. An augmented matrix includes the coefficients of the variables and constants from the system of equations without the variables. In this case, it becomes:\[ \begin{bmatrix} 1 & 1 & 1 & 0 \2 & 3 & 1 & 0 \ 3 & 5 & 1 & 0 \\end{bmatrix}\]

Step 2: Apply Gauss-Jordan Elimination

Perform row operations to get the matrix in row-reduced echelon form. Swap first and third row, then subtract the first row multiplied by 2 from the second row and the first row multiplied by 3 from the new third row. We get:\[ \begin{bmatrix} 3 & 5 & 1 & 0 \0 & 1 & -1 & 0 \ 0 & -1 & 1 & 0 \\end{bmatrix}\]Then, add the third row to the second row:\[ \begin{bmatrix} 3 & 5 & 1 & 0 \0 & 0 & 0 & 0 \ 0 & -1 & 1 & 0 \\end{bmatrix}\]Then, swap second and third row, and multiply the new second row by -1:\[ \begin{bmatrix} 3 & 5 & 1 & 0 \0 & 1 & -1 & 0 \0 & 0 & 0 & 0 \\end{bmatrix}\]And finally, divide the first row by 3:\[ \begin{bmatrix} 1 & 5/3 & 1/3 & 0 \0 & 1 & -1 & 0 \0 & 0 & 0 & 0 \\end{bmatrix}\]And subtract the first row multipled by 5/3 from the second row:\[ \begin{bmatrix} 1 & 0 & 0 & 0 \0 & 1 & -1 & 0 \0 & 0 & 0 & 0 \\end{bmatrix}\]

Step 3: Find the Solution

From the final matrix, write the corresponding system of linear equations, which gives \[\begin{align*}x &= 0 \y - z &= 0 \0 &= 0\end{align*}\]The third equation suggests that z is a free variable (can take any real number as its value). So, if we let \(z = s\) (with \(s\) being any real number), then \(y = z = s\), and \(x = 0\). So the system has infinitely many solutions.

Key Concepts

Gaussian elimination

To dive into solving systems of linear equations, Gaussian elimination is a widely used method. It involves transforming a matrix into a simpler form, making it easier to solve. With Gaussian elimination, we aim to turn our matrix into an upper triangular form, where all elements below the main diagonal are zeros.

This process is accomplished through a variety of row operations such as:

• Swapping two rows
• Multiplying a row by a non-zero constant
• Adding or subtracting a multiple of one row from another

By systematically applying these operations to an augmented matrix, we can streamline the system it's based on. Once in an upper triangular form, the values of the variables can be found using back-substitution. This step-by-step approach ensures a logical and effective pathway to solve the given equations.

row-reduced echelon form

Moving further, we encounter the row-reduced echelon form (RREF), a refined version of an already echelon-transformed matrix. Achieving this form makes analyzing solutions straightforward. Each leading coefficient is '1', and they move diagonally from the upper left to the bottom right, with zeros elsewhere.

The path to RREF involves continuous application of row operations, ensuring:

• Each pivot element is the only non-zero entry in its column
• All pivots (leading ones) are to the left of any pivots in rows below
• Rows of all zero are at the bottom

Completing this form helps identify whether we have a unique solution, no solution, or infinitely many solutions, as in matrices having a row of zeros indicates dependencies among equations. Reaching RREF is often a vital step, especially when faced with large, complex systems.

system of linear equations

In mathematics, a system of linear equations is a collection of two or more linear equations involving the same set of variables. Solving such a system involves finding the values of the variables that satisfy all equations concurrently.

Linear systems can be:

• Consistent (having at least one solution)
• Inconsistent (having no solution)

The consistency and nature of solutions greatly depend on the matrix derived from the system.

In the context of Gaussian elimination, understanding the balance of these equations is crucial. We use techniques like Gaussian or Gauss-Jordan elimination to discern the solution set, be it one unique solution, infinite solutions (the case of free variables), or no solution at all. Recognizing the type of solution helps us understand the relationship between our variables and the conditions they need to meet.