Problem 31 - Precalculus Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$\left\\{\begin{array}{rr} x+y-z= & 0 \\ x+2 y-3 z= & -3 \\ 2 x+3 y-4 z= & -3 \end{array}\right.$$

Step-by-Step Solution

Answer

The system is consistent with a general solution: \( x = 3 - 3t, y = -3 + 2t, z = t \).

Step 1: Write the Augmented Matrix

The given system can be written as an augmented matrix: \[ \begin{bmatrix} 1 & 1 & -1 & | & 0 \ 1 & 2 & -3 & | & -3 \ 2 & 3 & -4 & | & -3 \end{bmatrix} \]

Step 2: Perform Row Reduction

Perform row operations to transform the matrix into row-echelon form. We will subtract Row 1 from Row 2:\( R_2 = R_2 - R_1 \)This gives:\[ \begin{bmatrix} 1 & 1 & -1 & | & 0 \ 0 & 1 & -2 & | & -3 \ 2 & 3 & -4 & | & -3 \end{bmatrix} \]

Step 3: Eliminate First Column Elements

Eliminate the element in the first column of Row 3 by substituting Row 1:\( R_3 = R_3 - 2R_1 \)This results in:\[ \begin{bmatrix} 1 & 1 & -1 & | & 0 \ 0 & 1 & -2 & | & -3 \ 0 & 1 & -2 & | & -3 \end{bmatrix} \]

Step 4: Examine the Matrix for Consistency

Subtract Row 2 from Row 3:\( R_3 = R_3 - R_2 \)This gives:\[ \begin{bmatrix} 1 & 1 & -1 & | & 0 \ 0 & 1 & -2 & | & -3 \ 0 & 0 & 0 & | & 0 \end{bmatrix} \]Since the third row becomes all zeros, there is no contradiction, and the system is consistent.

Step 5: Write the Basic Solution

Next, deduce the solutions from the row-echelon form:1. From Row 2: \( y - 2z = -3 \)2. From Row 1: \( x + y - z = 0 \)Assume \( z = t \), a free variable. Then,\( y = -3 + 2t \) and \( x = 3 - 3t \).

Step 6: Write the General Solution

The general solution for the system with free variable \( z = t \) is:\[\begin{align*}x &= 3 - 3t, \y &= -3 + 2t, \z &= t.\end{align*}\]

Key Concepts

Augmented Matrix

An augmented matrix provides a compact way to represent a system of linear equations. It combines the coefficients of the variables with the constants of the equations into a single matrix format. In our example, we are given a system of three equations:

• Equation 1: \( x + y - z = 0 \)
• Equation 2: \( x + 2y - 3z = -3 \)
• Equation 3: \( 2x + 3y - 4z = -3 \)

To write the augmented matrix, each equation's coefficients form a row, and the constants form a separate column with a dividing line. Here, the matrix looks like this:\[\begin{bmatrix}1 & 1 & -1 & | & 0 \1 & 2 & -3 & | & -3 \2 & 3 & -4 & | & -3\end{bmatrix}\]The left side of the matrix contains the coefficients. The right column contains the constants from the equations. This form is useful for computational methods, as it allows for the application of matrix row operations to solve systems.

Row Reduction

Row reduction is a method used to solve linear systems by simplifying the augmented matrix using basic row operations. These operations include swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting scalar multiples of rows from each other. For our example, we aim to transform the matrix into simpler forms that reveal solutions.Performing Row Operations
Starting with the augmented matrix, we first subtract Row 1 from Row 2 to update Row 2:\[\begin{bmatrix}1 & 1 & -1 & | & 0 \0 & 1 & -2 & | & -3 \2 & 3 & -4 & | & -3\end{bmatrix}\]Next, eliminate the leading coefficient of the first element in Row 3 by subtracting twice Row 1 from Row 3:\[\begin{bmatrix}1 & 1 & -1 & | & 0 \0 & 1 & -2 & | & -3 \0 & 1 & -2 & | & -3\end{bmatrix}\]Finally, make Row 3 a row of zeros by subtracting Row 2 from Row 3:\[\begin{bmatrix}1 & 1 & -1 & | & 0 \0 & 1 & -2 & | & -3 \0 & 0 & 0 & | & 0\end{bmatrix}\]The presence of a row of zeros that doesn't create a contradiction indicates a consistent system, meaning it has at least one solution.

Row-Echelon Form

Row-echelon form is a simplified version of a matrix that helps to easily identify solutions to a linear system. A matrix in row-echelon form has the following characteristics:

• All nonzero rows are above rows of zeros.
• The leading coefficient (also known as the pivot) of a nonzero row is always to the right of the leading coefficient of the row above it.
• The leading coefficient is typically 1.

In our solved example, the matrix has been reduced to row-echelon form:\[\begin{bmatrix}1 & 1 & -1 & | & 0 \0 & 1 & -2 & | & -3 \0 & 0 & 0 & | & 0\end{bmatrix}\]Extracting Solutions
From this form, you can easily deduce the solutions. Each nonzero row represents an equation:

• Equation from Row 1: \( x + y - z = 0 \)
• Equation from Row 2: \( y - 2z = -3 \)

Row 3 suggests there are infinitely many solutions because it's all zeros, leaving variables like \( z \) free to take any value. This implies that other variables \( x \) and \( y \) are dependent on \( z \), leading to the general solution:

• \( x = 3 - 3t \)
• \( y = -3 + 2t \)
• \( z = t \)

Where \( t \) is any real number.