Problem 67 - 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. $$ \left\\{\begin{array}{r} x+y-5 z=3 \\ x-2 z=1 \\ 2 x-y-z=1 \end{array}\right. $$

Step-by-Step Solution

Answer

The given system of equations has no solution.

Step 1: Formulate the Augmented Matrix

From the given system of equations, the augmented matrix is: \[\begin{bmatrix} 1 & 1 & -5 & 3 \\ 1 & 0 & -2 & 1 \\ 2 & -1 & -1 & 1 \end{bmatrix}\]

Step 2: Apply Gaussian Elimination

Perform row operations to bring the matrix in echelon form. Subtract the first row from the second row and subtract twice the first row from the third row. The resulting matrix is: \[\begin{bmatrix} 1 & 1 & -5 & 3 \\ 0 & -1 & 3 & -2 \\ 0 & -3 & 9 & -5 \end{bmatrix}\] Next, multiply the second row by (-1) and add thrice the new second row to the third row. We obtain: \[\begin{bmatrix} 1 & 1 & -5 & 3 \\ 0 & 1 & -3 & 2 \\ 0 & 0 & 0 & -1 \end{bmatrix}\]

Step 3: Back Substitution

From the last row, we have the equation \(0=x\), which is a contradiction, indicating the system of equations has no solution.

Key Concepts

Gaussian elimination

Gaussian elimination is a method for solving a system of linear equations. It transforms the system's augmented matrix into a simpler form called row-echelon form, which makes it easier to find the solutions. This process involves performing a series of row operations to achieve zeros below the leading coefficients (also known as pivots) in each column. There are three types of row operations you can perform:

• Swapping two rows.
• Multiplying a row by a nonzero constant.
• Adding or subtracting a multiple of one row to another row.

The goal is to have the augmented matrix transform into a shape like this: \[\begin{bmatrix} 1 & a_{12} & a_{13} & \cdots & b_1 \ 0 & 1 & a_{23} & \cdots & b_2 \ 0 & 0 & 1 & \cdots & b_3 \ \end{bmatrix}\]This method simplifies solving the system to back substitution, provided a solution exists.

augmented matrix

An augmented matrix is a way of representing a system of linear equations. By using a single matrix, we can simultaneously work with the coefficients of the system as well as the constant terms. This combination helps simplify the solution process using methods like Gaussian elimination. For example, the system of equations:\( x + y - 5z = 3 \)\( x - 2z = 1 \)\( 2x - y - z = 1 \)can be written as the augmented matrix:\[\begin{bmatrix}1 & 1 & -5 & 3 \1 & 0 & -2 & 1 \2 & -1 & -1 & 1\end{bmatrix}\]In this matrix, the rows correspond to each equation, the columns on the left relate to the variables’ coefficients, and the final column contains the constant terms. The augmented matrix representation provides a clear and structured method to manage and manipulate the system's data when seeking solutions.

system of linear equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. The goal is to find values for these variables that satisfy all the equations simultaneously. Each equation in the system relates the variables with constants, typically in the form of \(ax + by + cz + \,\cdots\, = d\).Such systems can either have

• a unique solution,
• infinitely many solutions, or
• no solution at all.

Analyzing the solutions requires graphing the equations, using matrices, or substitution methods. In the case of the exercise at hand, Gaussian elimination is used to determine that this specific system has no solution, as the row reduction process yields a contradiction (\(0 = -1\)), showing inconsistency in the equations.

back substitution

Back substitution is a method used to solve a system of linear equations after it has been transformed into an upper triangular form using Gaussian elimination. Once the system is in row-echelon form, you can solve for variables starting from the bottom row upwards. For example, if you're left with something like:\[\begin{bmatrix}1 & a & b & c \0 & 1 & d & e \0 & 0 & 1 & f\end{bmatrix}\]The equation corresponding to the last row is \( z = f \). Substituting \( z \) into the second-to-last row provides a solution for \( y \ (y = e - dz)\). Continue working upwards to solve for \( x \) using previously calculated values. Back substitution is critical as it allows you to concretely determine the values of all variables. However, in the exercise example, the system is found to be inconsistent prior to reaching this step; hence, a solution could not be obtained due to a contradictory equation.