Problem 23 - Precalculus: Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. \(\left\\{\begin{array}{r}x-4 z=1 \\ 2 x-y-6 z=4 \\ 2 x+3 y-2 z=8\end{array}\right.\)

Step-by-Step Solution

Answer

The system is consistent and the solution is \( x = 5, y = 0, z = 1 \).

Step 1: Convert System into Matrix Form

First, convert the system of equations into an augmented matrix. Using the given equations: 1) \( x - 4z = 1 \) 2) \( 2x - y - 6z = 4 \) 3) \( 2x + 3y - 2z = 8 \) The augmented matrix is: \[\begin{bmatrix}1 & 0 & -4 & | & 1 \2 & -1 & -6 & | & 4 \2 & 3 & -2 & | & 8 \end{bmatrix}\]

Step 2: Perform Row Operations to Simplify Matrix

Start simplifying the matrix using elementary row operations to create zeros below the pivot in the first column (first row, first column). Subtract Row 1 from Row 2 and Row 3: - For Row 2: \( R_2 - 2R_1 \rightarrow R_2 \)- For Row 3: \( R_3 - 2R_1 \rightarrow R_3 \)The matrix becomes: \[\begin{bmatrix}1 & 0 & -4 & | & 1 \0 & -1 & 2 & | & 2 \0 & 3 & 6 & | & 6 \end{bmatrix}\]

Step 3: Create Zero Below the Pivot in Second Column

Now, eliminate the third element in the second column by performing the operation:\( R_3 + 3R_2 \rightarrow R_3 \)The augmented matrix becomes:\[\begin{bmatrix}1 & 0 & -4 & | & 1 \0 & -1 & 2 & | & 2 \0 & 0 & 12 & | & 12 \end{bmatrix}\]

Step 4: Back Substitution

Now solve the matrix through back substitution:1. From the third row: \( 12z = 12 \) Solving for \( z \), we get \( z = 1 \).2. Substitute \( z = 1 \) in the second row: \( -y + 2(1) = 2 \) or \( -y + 2 = 2 \) Simplify to find \( y = 0 \).3. Substitute \( z = 1 \) in the first row: \( x - 4 \times 1 = 1 \) or \( x - 4 = 1 \) Simplify to find \( x = 5 \).

Step 5: Write the Solution

The solution to the system of equations:\[ x = 5, \quad y = 0, \quad z = 1 \]. These values satisfy all three given equations in the system.

Key Concepts

Understanding Augmented Matrices

An augmented matrix is a handy way to represent a system of linear equations as a matrix. It combines the coefficients of the variables on one side, with the constants from the equations on the other side. For example, in the original system, the equations are represented as: 1) \( x - 4z = 1 \) 2) \( 2x - y - 6z = 4 \) 3) \( 2x + 3y - 2z = 8 \). These equations are transformed into an augmented matrix: \[\begin{bmatrix}1 & 0 & -4 & | & 1 \ 2 & -1 & -6 & | & 4 \ 2 & 3 & -2 & | & 8 \end{bmatrix}\] The vertical line differentiates the matrix into two parts:

• The left part contains the coefficients of the variables.
• The right part comprises the constants from the equations.

This representation is crucial as it simplifies calculations and allows operations to be performed in a more structured format.

Exploring Elementary Row Operations

Elementary row operations are the tools we use to manipulate matrices in order to solve systems of linear equations. These operations are essential in simplifying matrices, serving as steps towards achieving a row-echelon form, or ultimately, a reduced row-echelon form. The three fundamental types of row operations are:

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

In our example, we've applied these operations to transform the initial matrix:- Subtracting Row 1 from Row 2 and Row 3 to create zeros below the pivot: - \( R_2 - 2R_1 \rightarrow R_2 \) - \( R_3 - 2R_1 \rightarrow R_3 \)This simplifies the augmented matrix to: \[\begin{bmatrix}1 & 0 & -4 & | & 1 \ 0 & -1 & 2 & | & 2 \ 0 & 3 & 6 & | & 6 \end{bmatrix}\]Continuing with this method, we ultimately simplify the matrix to prepare it for back substitution.

Decoding Back Substitution

Back substitution is the final step in solving a linear system once the matrix has been simplified into an upper triangular form. This method involves solving the equations from the bottom up to find the values of the unknowns. Consider the simplified matrix:\[\begin{bmatrix}1 & 0 & -4 & | & 1 \ 0 & -1 & 2 & | & 2 \ 0 & 0 & 12 & | & 12 \end{bmatrix}\] - Start with the last row, which gives us \( 12z = 12 \). Solving this, we find \( z = 1 \).- Substitute \( z = 1 \) back into the second row: \( -y + 2z = 2 \) becomes \( -y + 2 = 2 \). This simplifies to \( y = 0 \).- Finally, substitute \( z = 1 \) and \( y = 0 \) into the first row: \( x - 4z = 1 \) turns into \( x - 4 = 1 \), yielding \( x = 5 \).The values \( x = 5 \), \( y = 0 \), and \( z = 1 \) satisfy the system, demonstrating the effectiveness of back substitution in solving linear equations efficiently.