Problem 21 - Precalculus Mathematics for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{array}{r} x+y+z=2 \\ 2 x-3 y+2 z=4 \\ 4 x+y-3 z=1 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (1, 0, 1)\).

Step 1: Write the Augmented Matrix

Start by writing the augmented matrix for the system of equations: \[\begin{bmatrix}1 & 1 & 1 & | & 2 \2 & -3 & 2 & | & 4 \4 & 1 & -3 & | & 1 \\end{bmatrix} \]

Step 2: Perform Row Operations to Get Upper Triangular Form

Use row operations to form an upper triangular matrix. Subtract 2 times the first row from the second row and 4 times the first row from the third row:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & -5 & 0 & | & 0 \0 & -3 & -7 & | & -7 \\end{bmatrix}\]

Step 3: Continue Row Operations to Isolate Leading Coefficients

Subtract \(\frac{3}{5}\) of the second row from the third row to eliminate the \(-3\) under the \(-5\):\[\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & -5 & 0 & | & 0 \0 & 0 & -7 & | & -7 \\end{bmatrix}\]

Step 4: Back Substitution to Find Variables

With the matrix in upper triangular form, perform back substitution. Start with the last row, which gives \(-7z = -7\), solving for \(z\) gives \(z = 1\). Substitute back to find other variables. From the second row: \(-5y = 0\), thus \(y = 0\). Substitute these into the first equation to find \(x + y + z = 2\), thus \(x = 1\).

Step 5: Verify the Solution

Substitute \(x = 1\), \(y = 0\), and \(z = 1\) back into the original equations to verify:- First equation: \(1 + 0 + 1 = 2\), correct.- Second equation: \(2(1) - 3(0) + 2(1) = 4\), correct.- Third equation: \(4(1) + 0 + -3(1) = 1\), correct.All solutions satisfy the original equations.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that depict a straight line when plotted on a graph. They are usually expressed in the form \( ax + by + cz = d \), where \( a, b, c \) are coefficients and \( x, y, z \) are variables. In the given exercise, you have three linear equations, each representing a plane in three-dimensional space. The intersection of these planes, if unique, corresponds to the solution of the system.
For example, consider the equations:

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

Each equation can be visualized as a distinct plane. The task is to find a single point (if it exists) where all these planes intersect, which represents the unique solution of the system.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It combines the coefficients and the constants from the equations into one matrix. This matrix serves as a basis for performing operations to solve the equations.
For the given system of equations, the augmented matrix looks like:
\[\begin{bmatrix} 1 & 1 & 1 & | & 2\2 & -3 & 2 & | & 4\4 & 1 & -3 & | & 1 \end{bmatrix} \] The first three columns represent the coefficients of the variables \( x, y, z \) while the last column, separated by a vertical line, contains the constants from each equation. This matrix serves as a starting point for applying Gaussian elimination, helping to systematize the solution process.

Row Operations

Row operations are essential in transforming an augmented matrix into a row-echelon form or reduced row-echelon form during Gaussian elimination. They include: swapping two rows, multiplying a row by a non-zero scalar, and adding or subtracting the multiple of one row to another row. These operations help in simplifying the matrix and isolating the coefficients of variables.
In the solution process:

• First, you subtract 2 times the first row from the second row to eliminate the \( 2x \) in the second row, and 4 times the first row from the third row to eliminate \( 4x \).
• Then, subtract \( \frac{3}{5} \) of the resulting second row from the third row to eliminate the term below the \(-5y \).

These steps modify the matrix into an upper triangular form, simplifying the next steps in finding the solution.

Back Substitution

Back substitution is the final step in solving a system of linear equations using Gaussian elimination. Once the matrix has been transformed into an upper triangular form, back substitution is used to find the values of the variables, starting from the last row upwards.
For the exercise:

• Start with the last row \( -7z = -7 \), which simplifies to \( z = 1 \).
• Move to the second row, \( -5y = 0 \), giving \( y = 0 \).
• Finally, substitute \( y = 0 \) and \( z = 1 \) back into the first row equation \( x + y + z = 2 \), leading to \( x = 1 \).

With these values, you solve the system and find the unique point where all planes intersect, fulfilling the conditions of each equation in the system.