Problem 68 - 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. $$ \left\\{\begin{array}{rr} x & -3 z= & -2 \\ 3 x+y-2 z & =5 \\ 2 x+2 y+z & =4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 40\), \(y = 53\), and \(z = -14\).

Step 1: Convert to Augmented Matrix

Rewrite the system of equations as an augmented matrix: \[ \left[\begin{array}{ccc|c} 1 & 0 & -3 & -2 \ 3 & 1 & -2 & 5 \ 2 & 2 & 1 & 4 \end{array}\right] \]Where the first column represents the coefficient of \(x\), the second column represents coefficient of \(y\), the third column represents coefficient of \(z\) and column after the line represents the constants on the right side of the equation.

Step 2: Apply Gauss-Jordan Elimination

Gauss-Jordan elimination involves using row operations to transform the initial matrix into a row-reduced echelon form. For this problem, first subtract 3 times the first row from the second row and 2 times the first row from the third row to eliminate \(x\) from the second and third rows. \[ \left[\begin{array}{ccc|c} 1 & 0 & -3 & -2 \ 0 & 1 & 3 & 11 \ 0 & 2 & 7 & 8 \end{array}\right] \] Then, subtract 2 times the second row from the third row to eliminate \(y\) from the third row: \[ \left[\begin{array}{ccc|c} 1 & 0 & -3 & -2 \ 0 & 1 & 3 & 11 \ 0 & 0 & 1 & -14 \end{array}\right] \]

Step 3: Determine Solutions

The row-reduced echelon form of the matrix correspond the equations: \(x - 3z = -2\), \(y + 3z = 11\), and \(z = -14\). Solving these equation for \(x\), \(y\), and \(z\) respectively provide the solutions: \(x = 40\), \(y = 53\), \(z = -14\)

Key Concepts

Gaussian Elimination

The Gaussian Elimination method is a systematic approach used to solve systems of linear equations. It involves employing a sequence of row operations to simplify the augmented matrix of a system until its upper triangular form is revealed. This form makes it easier to back substitute and find the values of the variables.

Key techniques include:

• Swapping two rows, if necessary.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting a multiple of one row to another row to eliminate variables.

The primary goal here is to get a matrix where the coefficients below the diagonal become zero, gradually simplifying the system. Once in upper triangular form, we can use back-substitution to solve for the unknowns starting from the bottommost equation.

Gauss-Jordan Elimination

Gauss-Jordan Elimination goes a step beyond Gaussian Elimination by aiming to reduce the matrix to its simplest possible form - the reduced row echelon form (RREF). This method simplifies the matrix further after achieving the upper triangular form by eliminating all coefficients above the diagonal as well.

This process is accomplished through continued row operations similar to those in Gaussian Elimination:

• Make leading coefficients (also known as pivots) equal to 1.
• Clear out all other entries in the pivot's column to make them zero.

The resulting RREF allows us to directly read off the variable values without back-substitution, as every variable corresponds to a unique equation in the system. This method is powerful for solving systems with complex dependencies.

Augmented Matrix

An augmented matrix is a convenient way of representing a system of linear equations. It combines the coefficients of the variables in the system and the constants from the equations into a single matrix.

For example, consider the system of equations:

• Equation 1: \(x - 3z = -2\)
• Equation 2: \(3x + y - 2z = 5\)
• Equation 3: \(2x + 2y + z = 4\)

The augmented matrix formed would be:\[ \left[\begin{array}{ccc|c}1 & 0 & -3 & -2 \3 & 1 & -2 & 5 \2 & 2 & 1 & 4 \\end{array}\right] \]The vertical line in the matrix differentiates the coefficients from the constants, helping to keep track of the equivalent equations during the elimination process.

Row Echelon Form

Row Echelon Form (REF) is a specific arrangement of a matrix where all non-zero rows appear first, and within each non-zero row, the leading entry (the first non-zero number from the left) is to the right of the leading entry in the row above it.

Characteristics of REF include:

• All rows with zero elements, if any, are at the bottom of the matrix.
• The leading entry in any non-zero row is always '1'.
• Each leading '1' is the only non-zero entry in its column.

Achieving a row echelon form is often the penultimate step in Gaussian Elimination. It sets the stage for back-substitution or can be further elaborated into reduced row echelon form through Gauss-Jordan elimination for direct solutions.