Problem 22 - Precalculus Mathematics for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{aligned} x+y+z &=4 \\ -x+2 y+3 z &=17 \\ 2 x-y &=-7 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The unique solution is \( x = -2 \), \( y = 3 \), \( z = 3 \).

Step 1: Write the Augmented Matrix

The given system of equations can be written in matrix form as an augmented matrix. This involves writing the coefficients and constants in a matrix format:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \-1 & 2 & 3 & | & 17 \2 & -1 & 0 & | & -7 \\end{bmatrix}\]

Step 2: Perform Row Operations for Triangular Form

First, eliminate the 'x' term from the second and third rows. Add Row 1 to Row 2 and multiply Row 1 by 2 and add it to Row 3:- \( R_2 = R_2 + R_1 \) gives \( [0, 3, 4, |, 21] \)- \( R_3 = R_3 - 2R_1 \) gives \( [0, -3, -2, |, -15] \)Now the matrix is:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \0 & 3 & 4 & | & 21 \0 & -3 & -2 & | & -15 \\end{bmatrix}\]

Step 3: Continue Elimination below Pivot

Add Row 2 to Row 3 to eliminate the 'y' term:- \( R_3 = R_3 + R_2 \) results in \( [0, 0, 2, |, 6] \)The matrix now becomes:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \0 & 3 & 4 & | & 21 \0 & 0 & 2 & | & 6 \\end{bmatrix}\]

Step 4: Solve for Leading Variable "z"

Use back substitution to solve for 'z':- From Row 3, \( 2z = 6 \), hence \( z = 3 \).

Step 5: Solve for "y" using Back Substitution

Substitute \( z = 3 \) back into Row 2:\( 3y + 4(3) = 21 \) gives \( 3y + 12 = 21 \).Solving for \( y \), subtract 12 from both sides to get \( 3y = 9 \), hence \( y = 3 \).

Step 6: Solve for "x" using Back Substitution

Substitute \( y = 3 \) and \( z = 3 \) back into Row 1:\( x + 3 + 3 = 4 \) yields \( x + 6 = 4 \).Solving for \( x \), subtract 6 from both sides to get \( x = -2 \).

Step 7: Check and Interpret Solution

The solution is \( x = -2 \), \( y = 3 \), \( z = 3 \). Check the solution by substituting these values back into the original equations to ensure they hold true. Each equation is satisfied, confirming the correctness of the solution.

Key Concepts

System of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. Each equation presents a relationship between the variables, expressed in the form of coefficients and constants. In our original exercise, we have a system of three equations in three variables: \(x, y,\) and \(z\). This system is:

• \(x + y + z = 4\)
• \(-x + 2y + 3z = 17\)
• \(2x - y = -7\)

The goal is to find the values of \(x, y,\) and \(z\) that satisfy all these equations simultaneously. Solving such systems can reveal a unique solution, no solution, or even infinitely many solutions. In this particular problem, the system has a unique solution. This means that there is exactly one set of values that meets all the criteria of each equation in the system. Understanding and solving systems of linear equations is a foundational concept in algebra and is essential for tackling more complex mathematical problems.

Augmented Matrix

The augmented matrix is a compact way of expressing a system of linear equations. This matrix includes both the coefficients of each variable and the constants from each equation. Creating an augmented matrix simplifies the process of performing operations to manipulate and solve the system. For our problem, we started with the following equations:

• \(x + y + z = 4\)
• \(-x + 2y + 3z = 17\)
• \(2x - y = -7\)

The augmented matrix derived from these equations is:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \-1 & 2 & 3 & | & 17 \2 & -1 & 0 & | & -7 \\end{bmatrix}\]In this matrix:

• The columns represent the coefficients for variables \(x\), \(y\), and \(z\).
• The last column after the bar (|) contains the constants from the right-hand side of each equation.

Operations such as row addition or scalar multiplication can be applied to this matrix to simplify the system into an easily solvable form, ultimately leading to a triangular matrix form where back substitution can be used to find the values of the unknowns.

Back Substitution

Once you have transformed the augmented matrix into a form where the equations are reduced to a triangular form, it's time to apply back substitution. Back substitution is a technique used to solve a system of equations from the "bottom up." This involves starting with the last equation, which should ideally contain one variable, solving for that variable, and then substituting that value back into the preceding equations.In our exercise:

• The last row of the transformed matrix yields \(2z = 6\), which simplifies to \(z = 3\).
• This \(z\) value is substituted back into the second row equation (\(3y + 4z = 21\)) to solve for \(y\). Using \(z = 3\), we find \(y = 3\).
• Finally, substituting both \(z = 3\) and \(y = 3\) into the first row equation (\(x + y + z = 4\)), we solve for \(x\), finding \(x = -2\).

This reverse approach leverages the already simplified equations and builds upon the previous calculations to determine each unknown, eventually yielding the unique solution for the entire system: \(x = -2\), \(y = 3\), \(z = 3\). Therefore, back substitution is a critical step in efficiently solving a system of linear equations by Gaussian elimination.