Problem 24 - 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. $$\left\\{\begin{aligned} 10 x+10 y-20 z &=60 \\ 15 x+20 y+30 z &=-25 \\\\-5 x+30 y-10 z &=45 \end{aligned}\right.$$

Step-by-Step Solution

Answer

\( x = \frac{184}{21}, y = -\frac{259}{21}, z = -\frac{22}{21} \)

Step 1: Write the augmented matrix

First, convert the system of equations into an augmented matrix form where each coefficient and constant term of the equations forms a row:\[\begin{bmatrix}10 & 10 & -20 & | & 60 \15 & 20 & 30 & | & -25 \-5 & 30 & -10 & | & 45\end{bmatrix}\]

Step 2: Normalize the first row

Divide the entire first row by the leading coefficient of 10 to make it 1 for easier elimination:\[\begin{bmatrix}1 & 1 & -2 & | & 6 \15 & 20 & 30 & | & -25 \-5 & 30 & -10 & | & 45\end{bmatrix}\]

Step 3: Eliminate the first column

Use row operations to eliminate the coefficients below the first element in the first column. Subtract 15 times the first row from the second row, and add 5 times the first row to the third row:\[\begin{bmatrix}1 & 1 & -2 & | & 6 \0 & 5 & 60 & | & -115 \0 & 35 & 0 & | & 75\end{bmatrix}\]

Step 4: Normalize the second row

Divide the second row by 5 to simplify the row, which helps in the next elimination step:\[\begin{bmatrix}1 & 1 & -2 & | & 6 \0 & 1 & 12 & | & -23 \0 & 35 & 0 & | & 75\end{bmatrix}\]

Step 5: Eliminate the second column

Use the second row to eliminate the coefficient in the second column of the third row by subtracting 35 times the second row from the third row:\[\begin{bmatrix}1 & 1 & -2 & | & 6 \0 & 1 & 12 & | & -23 \0 & 0 & -420 & | & 880\end{bmatrix}\]

Step 6: Solve for third variable (z)

With the third row, solve for the variable \(z\). Divide the entire row by -420:\[\begin{bmatrix}1 & 1 & -2 & | & 6 \0 & 1 & 12 & | & -23 \0 & 0 & 1 & | & -rac{22}{21}\end{bmatrix}\]

Step 7: Back-substitute to find y

Substitute \(z = -\frac{22}{21}\) into the second row to solve for \(y\):\( y + 12(-\frac{22}{21}) = -23 \)Simplify to find \( y = -\frac{259}{21} \).

Step 8: Back-substitute to find x

Substitute \(y\) and \(z\) into the first row to solve for \(x\):\( x + (-\frac{259}{21}) - 2(-\frac{22}{21}) = 6 \)Simplify to find \( x = \frac{184}{21} \).

Step 9: Final Step: Write the solution

The unique solution to the linear system is \( x = \frac{184}{21} \), \( y = -\frac{259}{21} \), and \( z = -\frac{22}{21} \).

Key Concepts

Gauss-Jordan Elimination

Gauss-Jordan elimination is a refined method of Gaussian elimination used to solve systems of linear equations, eventually leading to the identity matrix on the left side of the augmented matrix. The process involves using row operations to reach a state where the augmented matrix reflects the variables' coefficients, allowing an immediate read-off of their values. This method is particularly helpful because it doesn't just find one value at a time but systematically simplifies the entire array of equations.

The ultimate goal is to create a matrix that has 1's on the diagonal from top left to bottom right and 0's elsewhere in the left segment of the augmented matrix. This matrix transformation shows the solution to the system, making Gauss-Jordan elimination a powerful solution technique.

Linear Equations

Linear equations are algebraic expressions that represent a straight line when graphed. These equations usually take the form of:

\[ ax + by + cz = d \]
Here, \( x \), \( y \), and \( z \) represent the variables, while \( a \), \( b \), and \( c \) are their respective coefficients, and \( d \) is a constant. The system given in the exercise showcases three linear equations, which together form a system that can be solved using methods like Gauss-Jordan elimination.

Such systems of linear equations can have one solution, no solution, or infinitely many solutions depending on the nature of the equations and their relationships with each other.

Augmented Matrix

An augmented matrix is a streamlined way to represent a system of linear equations. The matrix comprises the coefficients of the variables and the constants from the equations, presented in a tidy structured layout. For instance, the system given is transformed into an augmented matrix as follows:

\[\begin{bmatrix}10 & 10 & -20 & | & 60 \15 & 20 & 30 & | & -25 \-5 & 30 & -10 & | & 45\end{bmatrix}\]
The vertical line separates the coefficients on the left from the constants on the right, making it visually clearer to apply operations in methods like Gaussian or Gauss-Jordan elimination.

Row Operations

Row operations are key tools in solving systems of linear equations via a matrix representation. They can alter the rows of an augmented matrix in specific ways, without changing the solution set. There are three primary types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting the multiple of one row to or from another row.

Using these row operations, we can progressively simplify the matrix. For example, making a leading coefficient one, eliminating variables below leading ones, and so on, until the system is easily solvable. Each type of row operation aids in zeroing out elements or adjusting values to resolve the variables step-by-step as if peeling layers of an onion, ultimately revealing the core solution.