Problem 24 - Precalculus: Mathematical for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{array}{rr} 10 x+10 y-20 z= & 60 \\ 15 x+20 y+30 z= & -25 \\ -5 x+30 y-10 z= & 45 \end{array}\right.$$

Step-by-Step Solution

Answer

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

Step 1: Write the Augmented Matrix

Start by converting the system of equations into an augmented matrix. Each row represents an equation and each column represents the coefficients of one of the variables or the constants on the right-hand side:\[ \begin{bmatrix} 10 & 10 & -20 & | & 60 \ 15 & 20 & 30 & | & -25 \ -5 & 30 & -10 & | & 45 \end{bmatrix} \].

Step 2: Row Reduction - Eliminate x from 2nd and 3rd Rows

Perform row operations to eliminate the variable \( x \) from the second and third rows. First, modify the second row by subtracting \( \frac{3}{2} \times \) the first row from it. Then, add the first row to the third row.After performing operations:\[ \begin{bmatrix} 10 & 10 & -20 & | & 60 \ 0 & 5 & 60 & | & -115 \ 0 & 40 & -30 & | & 105 \end{bmatrix} \].

Step 3: Simplify Second Row

The current second row can be simplified by dividing all the elements by 5:\[ \begin{bmatrix} 10 & 10 & -20 & | & 60 \ 0 & 1 & 12 & | & -23 \ 0 & 40 & -30 & | & 105 \end{bmatrix} \].

Step 4: Row Reduction - Eliminate y from Third Row

To eliminate \( y \) from the third row, subtract 40 times the second row from the third row:\[ \begin{bmatrix} 10 & 10 & -20 & | & 60 \ 0 & 1 & 12 & | & -23 \ 0 & 0 & -510 & | & 1025 \end{bmatrix} \].

Step 5: Solve for z

From the third row, we can solve for \( z \) by dividing through by -510:\[ z = -2 \].

Step 6: Back Substitution - Solve for y

Use the second row to find \( y \) by substituting \( z = -2 \):\[ 0 + 1 \cdot y + 12 \cdot (-2) = -23 \]\[ y = -23 + 24 = 1 \].

Step 7: Back Substitution - Solve for x

Use the first row to find \( x \) by substituting \( y = 1 \) and \( z = -2 \):\[ 10x + 10 \cdot 1 - 20 \cdot (-2) = 60 \]\[ 10x + 10 + 40 = 60 \]\[ 10x = 10 \]\[ x = 1 \].

Step 8: Verify the Solution

Substitute \( x = 1 \), \( y = 1 \), and \( z = -2 \) back into the original equations to verify their correctness. All equations should be satisfied for the found values.

Key Concepts

Linear Equations

Linear equations are mathematical expressions consisting of variables raised to the first power and globally represent a straight line when graphed. They can have one or more variables, often denoted as \( x, y, \) or \( z \). In a system of linear equations, multiple equations exist simultaneously, and the goal is often to find the values of the variables that satisfy all equations.

• Each equation balances a combination of variables and constants on either side of the "equal to" sign.
• In solving systems of linear equations, methods like substitution, elimination, and matrix operations are commonly used.
• The unique solution represents the intersection point of all involved equations' lines in a two-dimensional plane or hyperplanes in higher dimensions.

Understanding linear equations is crucial as they form the foundation for much more complex algebraic concepts and applications in fields such as engineering, physics, and economics.

Augmented Matrix

An augmented matrix is an essential tool used to represent a system of linear equations. It simplifies working with multiple equations by consolidating them into a matrix format. This matrix includes the coefficients of variables and the constants from the equations on the right side.

• The columns of the augmented matrix correspond to different variables and the constant terms.
• The row represents each individual equation from the system.
• For example, consider a system provided in the exercise:\[\begin{bmatrix} 10 & 10 & -20 & | & 60 \ 15 & 20 & 30 & | & -25 \ -5 & 30 & -10 & | & 45 \end{bmatrix}\]This matrix includes coefficients and constants from the given equations.

The advantage of using an augmented matrix is that it provides a straightforward approach to apply matrix operations for finding solutions, especially during Gaussian elimination.

Row Reduction

Row reduction is a method used to solve systems of linear equations transformed into an augmented matrix. It involves executing a series of operations on the rows of a matrix to simplify it, making finding solutions easier.

• The primary operations include:
  • Swapping two rows
  • Multiplying a row by a non-zero constant
  • Adding or subtracting a multiple of one row from another row
• The goal is to convert the matrix into an equivalent form, often triangular or row-echelon form.
• In the example given, the exercise uses row reduction to simplify the augmented matrix and solve for each variable systematically.By progressively simplifying the matrix, finding solutions for \( x, y, \) and \( z \) is achieved efficiently.

Gauss-Jordan Elimination

The Gauss-Jordan elimination method is an extension of Gaussian elimination. It further simplifies the augmented matrix to reach the reduced row-echelon form, facilitating direct reading of the solution.

• Unlike Gaussian elimination, Gauss-Jordan goes beyond to make leading coefficients into one and ensures all entries above and below leading ones are zero.
• In essence, it transforms the matrix so that it directly reveals the solutions for the variables.
• For instance, in the original exercise solution:
  • Initial elimination and simplification steps help gradually simplify the matrix.
  • The row operations applied for each variable ensure a straightforward back-substitution where the results can be directly read off as \( x = 1 \), \( y = 1 \), and \( z = -2 \).
• This method is highly effective for solutions of systems of equations, especially those requiring precision, since it eliminates the errors introduced in step-by-step back-substitution.Understanding such elimination processes empowers students to tackle complex systems with confidence.