Problem 75 - 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. $$ \left\\{\begin{array}{r} x+2 y=0 \\ -x-y=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system is \(x = 0\) and \(y = 0\).

Step 1: Converting the system to matrix form

We write the system in the form AX = B, where A is the matrix of coefficients, X is the column matrix \[ \begin{bmatrix}x \ y \end{bmatrix} \], and B is the matrix on the right-hand side of the equation. Therefore, \[ \begin{bmatrix}1 & 2 \ -1 & -1 \end{bmatrix} \begin{bmatrix}x \ y \end{bmatrix} = \begin{bmatrix}0 \ 0 \end{bmatrix} \].

Step 2: Applying Gaussian elimination

We can use the augmented matrix to apply Gaussian elimination. The augmented matrix is \[ \begin{bmatrix}1 & 2 & 0 \ -1 & -1 & 0 \end{bmatrix} \]. We then add Row 1 to Row 2 to form a new Row 2, \[ \begin{bmatrix}1 & 2 & 0 \ 0 & 1 & 0 \end{bmatrix} \].

Step 3: Applying back-substitution

The above matrix shows our system of equations as \(x + 2y = 0\) and \(y = 0\). Using back-substitution, substitute \(y = 0\) into the first equation to get \(x = 0\).

Key Concepts

Gaussian Elimination

To understand Gaussian elimination, picture a system of equations as a balancing act where we aim to find the weight (value) of each element (variable) that maintains the balance (equation). In simpler terms, it's a method used to solve systems of linear equations.

Gaussian elimination transforms these systems into a triangle-like form, which makes it easy to solve for the variables step by step. This transformation is achieved by performing row operations to construct an upper-triangular matrix of coefficients:

• Switching rows (if necessary),
• Multiplying a row by a non-zero scalar,
• Adding or subtracting rows from each other.

These actions, when applied to the matrix representation of the system, do not affect the solution; they merely simplify the way we look at the equations. The final goal is to reach a point where each equation has one less variable than the equation above it, leading us towards an easily solvable equation at the bottom.

Gauss-Jordan Elimination

Taking a step further, Gauss-Jordan elimination continues the process of Gaussian elimination beyond the point of creating an upper triangular matrix. Instead, it aims for a matrix in reduced row echelon form (RREF), which makes it even simpler to read off the solutions directly from the matrix.

It involves additional steps such as:

• Making sure the leading coefficient (the first non-zero number from the left, in a non-zero row) is 1,
• Making all the numbers above and below this leading 1 to be zeroes by adding or subtracting multiples of one row to another.

Because Gauss-Jordan elimination involves creating zeroes both above and below the leading 1s, it's generally considered a more thorough method of simplification compared to Gaussian elimination, which only requires zeroes below.

Matrix Representation of Systems

To solve our systems using either Gaussian or Gauss-Jordan elimination, we first need to represent them in matrix form. A system of linear equations can be expressed as the matrix equation AX = B, where:

• A is the coefficient matrix,
• X is the column matrix of variables,
• and B is the column matrix of constants.

By organizing our equations in this way, we turn the problem into a type of puzzle where the pieces are the rows of the matrix, and solving the system is equivalent to arranging these rows in a way that reveals the values of our variables in X.

Back-Substitution

Once we have applied Gaussian elimination to create an upper triangular matrix with no non-zero elements below the main diagonal, we can use back-substitution to find the solutions for our variables.

Starting from the bottom of the matrix, we solve the last equation to find the value of the last variable. Then, we substitute this value into the equation above it to find the second-to-last variable, and so on, moving upwards through the system.

• This stepwise approach reduces the complexity of the initial system of equations,
• It is only applicable once the coefficient matrix is in triangular form – it wouldn't make much sense otherwise, as the equations wouldn't be simplified enough to solve directly.