Problem 29 - Precalculus Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$\left\\{\begin{aligned} 2 x+3 y-z &=1 \\ x+2 y &=3 \\ x+3 y+z &=4 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system is inconsistent; there is no solution.

Step 1: Write the system in augmented matrix form

Start by writing the given system of equations as an augmented matrix. The system is:\[\begin{align*}2x + 3y - z &= 1 \x + 2y &= 3 \x + 3y + z &= 4\end{align*}\]The augmented matrix is:\[\begin{bmatrix}2 & 3 & -1 & | & 1 \1 & 2 & 0 & | & 3 \1 & 3 & 1 & | & 4\end{bmatrix}\]

Step 2: Perform row operations to get zeros below pivot in column 1

Subtract the second row from the first row to eliminate the first element of the second row:\[ R_2 = R_1 - R_2 \]The augmented matrix becomes:\[\begin{bmatrix}2 & 3 & -1 & | & 1 \0 & 1 & -1 & | & -2 \1 & 3 & 1 & | & 4\end{bmatrix}\]Next, subtract the first row divided by 2 from the third row to eliminate the first element of the third row:\[ R_3 = R_3 - \frac{1}{2}R_1 \]The matrix is updated to:\[\begin{bmatrix}2 & 3 & -1 & | & 1 \0 & 1 & -1 & | & -2 \0 & 1.5 & 1.5 & | & 3.5\end{bmatrix}\]

Step 3: Eliminate second row's pivot from third row

Subtract 1.5 times the second row from the third row to eliminate the first element below the second pivot:\[ R_3 = R_3 - 1.5R_2 \]The matrix simplifies to:\[\begin{bmatrix}2 & 3 & -1 & | & 1 \0 & 1 & -1 & | & -2 \0 & 0 & 3 & | & 6\end{bmatrix}\]

Step 4: Convert to row-echelon form

We can see that the third row has a leading coefficient. Divide the entire row by 3:\[ R_3 = \frac{1}{3}R_3 \]We get the row-echelon form:\[\begin{bmatrix}2 & 3 & -1 & | & 1 \0 & 1 & -1 & | & -2 \0 & 0 & 1 & | & 2\end{bmatrix}\]

Step 5: Back-substitution to find the solutions

From the final row of the matrix, we have:\[ z = 2 \]Substitute \( z = 2 \) into the second row:\[ y - z = -2 \Rightarrow y - 2 = -2 \Rightarrow y = 0 \]Now substitute \( y = 0 \) and \( z = 2 \) into the first row:\[ 2x + 3y - z = 1 \Rightarrow 2x + 0 - 2 = 1 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2} \]

Step 6: Verify the solution

Substitute \( x = \frac{3}{2}, y = 0, z = 2 \) back into the original equations to ensure they satisfy all three equations:1. \[ 2 \cdot \frac{3}{2} + 3 \cdot 0 - 2 = 1 \Rightarrow 3 - 2 = 1 \] (Correct)2. \[ \frac{3}{2} + 2 \cdot 0 = 3 \Rightarrow \frac{3}{2} \, (Incorrect) \] 3. \[ \frac{3}{2} + 3 \cdot 0 + 2 = 4 \Rightarrow \frac{3}{2} + 2 = 4 \] (Incorrect)Equations 2 and 3 show inconsistency, indicating no solution.

Key Concepts

Augmented Matrix

When solving systems of linear equations, we often start by translating these equations into an augmented matrix. An augmented matrix is essentially a compact way of representing a system of equations. It combines the coefficients of the variables and the constants found on the other side of the equal sign into a single rectangular array. For example, consider the system:

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

These equations can be represented in the augmented matrix form as: \[\begin{bmatrix} 2 & 3 & -1 & | & 1 \ 1 & 2 & 0 & | & 3 \ 1 & 3 & 1 & | & 4 \end{bmatrix}\] Here, each row corresponds to an equation from the system, and the vertical bar separates the coefficients from the constants. This matrix format simplifies manipulation and helps us apply row operations for a more streamlined solution process.

Row Operations

To transform an augmented matrix and find solutions, we perform specific actions known as row operations. These operations help simplify the matrix while preserving the solutions to the system of equations. There are three main types of row operations:

• Swapping two rows: This operation changes the order of the equations without altering their relationships.
• Multiplying a row by a non-zero scalar: This operation scales an equation to simplify calculations.
• Adding or subtracting rows: This helps eliminate variables, leading to a simpler system to solve.

In our example, we often use these operations to create zeros below pivot positions, making the matrix easier to work with. For instance, subtracting the second row from the first helps eliminate certain elements and reshape the matrix toward row-echelon form.

Row-Echelon Form

The row-echelon form is a simplified version of a matrix that helps us easily extract solutions from a system of equations. In this form, the matrix has a stepped appearance, with leading coefficients creating a sort of stair pattern from top-left to bottom-right. Row-echelon form has several important characteristics:

• Any zero rows are at the bottom of the matrix.
• The leading coefficient of a non-zero row is always to the right of the leading coefficient of the row above.
• All elements below a leading coefficient are zeros.

The row-echelon form allows us to simplify a system of equations to the point where back-substitution can be easily applied. For the given exercise, transforming the matrix to row-echelon form helped in visualizing the system's dependency on one another, exposing that the solution wasn't viable across all equations.

Back-Substitution

Back-substitution is a method used to solve a system of equations once the augmented matrix has been transformed into an upper triangular form, usually as row-echelon form. This technique involves solving for the last variable first and using this solution to find the other variables step by step. Let's break it down:

• Start with the last row of the matrix, which gives us a clear equation for the last variable.
• Substitute this variable's value into the equation above to solve for the next variable.
• Continue this process moving upward until all variable values have been determined.

However, in certain circumstances, like in our example, if our substitution reveals inconsistencies in the system (conflicting solutions from different equations), it indicates the system is inconsistent and has no solutions. Understanding this process helps students analyze the relationships among equations deeply.