Problem 42 - College Algebra

Question

For the following exercises, solve the system by Gaussian elimination. $$ \begin{array}{c} x+y=2 \\ x+z=1 \\ -y-z=-3 \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = 0\), \(y = 2\), \(z = 1\).

Step 1: Write the system as an augmented matrix

The given system of equations can be represented as an augmented matrix. Each row corresponds to one equation and each column to one variable or its constant term. The matrix is written as: \[ \begin{bmatrix} 1 & 1 & 0 & | & 2 \ 1 & 0 & 1 & | & 1 \ 0 & -1 & -1 & | & -3 \end{bmatrix} \]

Step 2: Perform row operations to achieve upper triangular form

We start by making the first column below the first row zero. Subtract the first row from the second row. The new second row becomes: \[ (1 - 1)\,x,\, (0 - 1)\,y,\, (1 - 0)\,z,\, (1 - 2) \rightarrow 0, -1, 1, -1 \] So the augmented matrix becomes: \[ \begin{bmatrix} 1 & 1 & 0 & | & 2 \ 0 & -1 & 1 & | & -1 \ 0 & -1 & -1 & | & -3 \end{bmatrix} \]

Step 3: Simplify the second and third rows

Add the second row to the third row to eliminate the second element in the third row. The third row becomes: \[ (0 + 0), (-1 + -1), (1 + -1), (-1 + -3) \rightarrow 0, -2, 0, -4 \] The augmented matrix is now: \[ \begin{bmatrix} 1 & 1 & 0 & | & 2 \ 0 & -1 & 1 & | & -1 \ 0 & -2 & 0 & | & -4 \end{bmatrix} \]

Step 4: Further reduce to simpler form

Divide the third row by -2 to simplify: \[ 0, 1, 0, 2 \] Then swap it with the second row for ease of calculation. The matrix updates to: \[ \begin{bmatrix} 1 & 1 & 0 & | & 2 \ 0 & 1 & 0 & | & 2 \ 0 & -1 & 1 & | & -1 \end{bmatrix} \]

Step 5: Clear Y-term from the first and third equations

Subtract the second row from the first to eliminate 'y'. The new first row becomes: \[ (1), (1 - 1), (0), (2 - 2) \rightarrow 1, 0, 0, 0 \] Add the second row to the third row. The new third row is: \[ (0), (-1 + 1), (1 + 0), (-1 + 2) \rightarrow 0, 0, 1, 1 \] The modified augmented matrix now is: \[ \begin{bmatrix} 1 & 0 & 0 & | & 0 \ 0 & 1 & 0 & | & 2 \ 0 & 0 & 1 & | & 1 \end{bmatrix} \]

Step 6: Extract the solution from the matrix

The matrix is now in reduced row-echelon form, corresponding to equations: 1. \(x = 0\) 2. \(y = 2\) 3. \(z = 1\) The solution to the system is \((x, y, z) = (0, 2, 1)\).

Key Concepts

System of Equations

A system of equations consists of two or more equations with the same set of variables. The goal is to find values for these variables that satisfy all of the equations simultaneously. Each equation represents a relationship involving the variables. For instance, in our given system with equations like \(x+y=2\), \(x+z=1\), and \(-y-z=-3\), every equation encapsulates a unique expression of how the variables \(x\), \(y\), and \(z\) relate to each other.
These systems can have various outcomes:

• One Solution: Means a unique set of values exists that satisfies all the equations. This is the case in our example.
• No Solution: If no possible set of values satisfies all the equations.
• Infinite Solutions: When there are infinitely many values for the variables that can satisfy the system, often occurring if the equations are dependent.

The Gaussian elimination method aims to transform such systems into a simpler form (row-echelon form), allowing us to deduce the solution easily.

Augmented Matrix

An augmented matrix is a compact tool used in linear algebra to represent and solve a system of linear equations. This structure combines the coefficients of the variables and the constants from each equation into a single matrix.
Consider the system from the exercise:

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

This system can be translated into an augmented matrix as follows:\[\begin{bmatrix} 1 & 1 & 0 & | & 2 \1 & 0 & 1 & | & 1 \0 & -1 & -1 & | & -3 \end{bmatrix}\]
Here:

• The left side of the vertical line contains the coefficients of variables \(x\), \(y\), and \(z\).
• The right side of the vertical line includes the constant terms.

By systematically applying row operations, we can transform this matrix to reveal the solutions.

Row Operations

Row operations are mathematical manipulations used on matrices to simplify systems of equations and solve them efficiently. When we apply Gaussian elimination, we mainly perform three types of row operations:

• Row Swapping: Interchanging two rows can simplify calculations. This was used towards the end of our exercise for ease of computation.
• Row Multiplication: Multiplying a row by a non-zero scalar helps in obtaining zeros or simplifying coefficients. In our solution, we applied this to simplify the third row.
• Row Addition/Subtraction: Adding or subtracting one row from another is crucial to eliminate variables, which is a frequent step in our solution to reach row echelon form.

By employing these operations, we transform our original augmented matrix into a simpler, row-echelon or reduced row-echelon form, ultimately isolating each variable and making the solution evident. Each operation maintains the original system's equivalence, ensuring the final solution is correct and valid.