Problem 32 - Precalculus: Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. \(\left\\{\begin{aligned} x-2 y-3 z &=5 \\ 2 x+y-z &=5 \\ 4 x-3 y-7 z &=5 \end{aligned}\right.\)

Step-by-Step Solution

Answer

The system is inconsistent and has no solution.

Step 1: Write the Augmented Matrix

Begin by writing the given system of equations as an augmented matrix. The system of equations is: \(\begin{aligned} x-2y-3z &= 5 \ 2x+y-z &= 5 \ 4x-3y-7z &= 5 \end{aligned}\). The corresponding augmented matrix is: \[\left[\begin{array}{ccc|c}1 & -2 & -3 & 5 \2 & 1 & -1 & 5 \4 & -3 & -7 & 5\end{array}\right]\]

Step 2: Perform Row Operations to Simplify

Perform row operations to convert the matrix to row-echelon form (REF). First, eliminate the first column below the first row. Use the first row to eliminate the \(x\)-components in the lower rows:- \(R_2 \to R_2 - 2 \cdot R_1\):\[\left[\begin{array}{ccc|c}1 & -2 & -3 & 5 \0 & 5 & 5 & -5 \4 & -3 & -7 & 5\end{array}\right]\]- \(R_3 \to R_3 - 4 \cdot R_1\):\[\left[\begin{array}{ccc|c}1 & -2 & -3 & 5 \0 & 5 & 5 & -5 \0 & 5 & 5 & -15\end{array}\right]\]

Step 3: Continue Row Reduction to REF

Next, reduce the second column. Eliminate the second row component in the third row:- \(R_3 \to R_3 - R_2\):\[\left[\begin{array}{ccc|c}1 & -2 & -3 & 5 \0 & 5 & 5 & -5 \0 & 0 & 0 & -10\end{array}\right]\]

Step 4: Check for Consistency

Examine the last row of the matrix. It represents the equation \(0x + 0y + 0z = -10\), which simplifies to \(0 = -10\). Since this is a contradiction, it implies there are no solutions to this system and the system is inconsistent.

Key Concepts

Augmented Matrix

To solve systems of linear equations, we often start by organizing our equations into an augmented matrix. An augmented matrix condenses the system into a grid of numbers representing the coefficients and constants from the equations. This makes it easier to visualize and manipulate the system for further steps. For the given equations:

• \( x - 2y - 3z = 5 \)
• \( 2x + y - z = 5 \)
• \( 4x - 3y - 7z = 5 \)

The corresponding augmented matrix is: \[\begin{bmatrix} 1 & -2 & -3 & | & 5 \ 2 & 1 & -1 & | & 5 \ 4 & -3 & -7 & | & 5 \end{bmatrix}\]Here, each row corresponds to an equation, and the vertical bar separates the coefficients from the constants. This setup allows us to use matrix operations to simplify the system.

Row Operations

Row operations are the tools we use to transform an augmented matrix into a simpler form. There are three types of row operations, which we can perform on the rows of the matrix:

• Swap two rows.
• Multiply a row by a nonzero constant.
• Add or subtract a multiple of one row to another row.

These operations preserve the solution set of the system. In our example, we apply these operations to eliminate variables from specific positions to approach a more easily solvable form. For instance, we used the first row to eliminate the \( x \)-components below it, and then further reduced the matrix by using the second row.

Row-Echelon Form

Row-echelon form (REF) is a simpler form of a matrix where

• All nonzero rows are above any rows of all zeros.
• The leading entry (or pivot) of each nonzero row is 1, located to the right of the leading coefficient of the row above.
• Zeros are below each pivot position within each pivot column.

By applying row operations as shown in the solution, we adjust the original matrix to meet these conditions. This makes it easier to observe the solutions or determine inconsistency in the system. Continuing from our row operations, we obtained a form revealing that the last row simplifies to a contradiction.

Inconsistent System

In a matrix, an inconsistent system arises when there's a row indicating a contradiction, such as 0 is equal to a non-zero number. This signals that the system of equations does not have any solutions. In our exercise, the final form of the matrix had the last row as \( \begin{bmatrix} 0 & 0 & 0 & | & -10 \end{bmatrix} \), implying the equation \( 0 = -10 \). Such a statement is logically impossible, leading to a conclusion that the system has no solutions. This particular case showcases the beauty of row techniques in revealing underlying relationships or contradictions within a system of equations.