Problem 15 - Algebra and Trigonometry

Question

15–24 The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{aligned} x-2 y+z &=1 \\ y+2 z &=5 \\ x+y+3 z &=8 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = 1\), \(y = 1\), and \(z = 2\).

Step 1: Write the Augmented Matrix

Convert the system of equations into an augmented matrix. The matrix becomes:\[\begin{bmatrix}1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \1 & 1 & 3 & | & 8\end{bmatrix}\]

Step 2: Eliminate x from the Second Row

Use the first row to eliminate the x-term (first position) in the second row. Since the current term is already 0, no row operations are needed.\[\begin{bmatrix}1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \1 & 1 & 3 & | & 8\end{bmatrix}\]

Step 3: Eliminate x from the Third Row

Eliminate the x-term from the third row by subtracting Row 1 from Row 3:\[R_3 = R_3 - R_1\]Resulting matrix:\[\begin{bmatrix}1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \0 & 3 & 2 & | & 7\end{bmatrix}\]

Step 4: Eliminate y from the Third Row

Use the second row to eliminate the y-term (second element) in the third row by substituting one half of the second row from the third row:\[R_3 = R_3 - 3 \times R_2\]Resulting matrix:\[\begin{bmatrix}1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \0 & 0 & -4 & | & -8\end{bmatrix}\]

Step 5: Solve for z

The third row now reads \(0z = -8\), which simplifies to \(-4z = -8\). Divide both sides by -4 to find \(z\):\[z = 2\]

Step 6: Back Substitution for y

Substitute \(z = 2\) back into the second row equation:\[0 + y + 2(2) = 5\]Simplify to find \(y\):\[y + 4 = 5 \y = 1\]

Step 7: Back Substitution for x

Substitute \(z = 2\) and \(y = 1\) back into the first row equation:\[x - 2(1) + 1(2) = 1\]Simplify to find \(x\):\[x - 2 + 2 = 1 \x = 1\]

Step 8: Present the Solution

The solution to the system of equations is \(x = 1\), \(y = 1\), and \(z = 2\). Final solution:\[(x, y, z) = (1, 1, 2)\]

Key Concepts

system of linear equations

A system of linear equations consists of multiple linear equations involving the same set of variables. The goal is to find a common solution that satisfies all the equations simultaneously. Such a system is expressed as a list of equations, for example: \[ \begin{aligned} x - 2y + z &= 1 \ y + 2z &= 5 \ x + y + 3z &= 8 \end{aligned} \].
Each equation in this system represents a flat surface in three-dimensional space, and the common intersection point of these surfaces gives us the solution with unique values for \(x\), \(y\), and \(z\).
There are various methods to solve such systems, but they all aim to simplify the equations to easily find the values of the variables involved.

augmented matrix

An augmented matrix is a simple method used to compactly represent a system of linear equations. It merges the coefficients of the variables and the constants from the equations into a single matrix form. For example, for our system of equations, the augmented matrix is:
\[ \begin{bmatrix} 1 & -2 & 1 & | & 1 \ 0 & 1 & 2 & | & 5 \ 1 & 1 & 3 & | & 8 \end{bmatrix} \].
Here, the coefficients of \(x\), \(y\), and \(z\) are aligned in the first three columns, while the constants from the equations are placed after a vertical bar.
This form simplifies the application of elimination methods, like Gaussian or Gauss-Jordan elimination.

Gauss-Jordan elimination

Gauss-Jordan elimination is an extension of Gaussian elimination. It involves performing a series of row operations to transform the augmented matrix into a form where finding the solutions is direct.
Let's revisit our matrix transformation process:
- We started with the matrix \( \begin{bmatrix} 1 & -2 & 1 & | & 1 \ 0 & 1 & 2 & | & 5 \ 1 & 1 & 3 & | & 8 \end{bmatrix} \).
- After some row operations, we aimed to achieve a matrix where each row represents a simpler equation with reduced variables.
The ultimate goal of Gauss-Jordan elimination is to convert the matrix to reduced row echelon form (RREF), leading directly to the variable solutions without further calculations.

back substitution

Back substitution is a technique used after simplifying a system of equations, making it ready for direct computation of variables' values. After using elimination methods, we ended up with a matrix in a simpler form:
\[ \begin{bmatrix} 1 & -2 & 1 & | & 1 \ 0 & 1 & 2 & | & 5 \ 0 & 0 & -4 & | & -8 \end{bmatrix} \].
With simplified equations, we calculated our unknowns, starting from the bottom:
- From the third row: \( -4z = -8 \), hence \( z = 2 \).
- Using \( z = 2 \) in the second row, \( y + 2(2) = 5 \) gives us \( y = 1 \).
- The values of \( y \) and \( z \) substituted back in the first row gives us \( x - 2(1) + 2 = 1 \), resulting in \( x = 1 \).
Back substitution confirms that we have found \( x = 1 \), \( y = 1 \), and \( z = 2 \), solving the system completely.