Problem 33 - Precalculus: Mathematics for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{array}{rr} x+2 y-z= & -2 \\ x & +z=0 \\ 2 x-y-z= & -3 \end{array}\right.$$

Step-by-Step Solution

Answer

The unique solution is \( x = -1 \), \( y = 0 \), and \( z = 1 \).

Step 1: Convert System to Augmented Matrix

A system of linear equations can be written in matrix form. The coefficients of each variable become elements of a matrix, and the constants on the right side of each equation become elements of the augmented column. The system \[ \begin{cases} x + 2y - z = -2 \ x + 0y + z = 0 \ 2x - y - z = -3 \end{cases} \] can be written as the augmented matrix: \[ \begin{bmatrix} 1 & 2 & -1 & | & -2 \ 1 & 0 & 1 & | & 0 \ 2 & -1 & -1 & | & -3 \end{bmatrix} \]

Step 2: Apply Row Operations to Achieve Row Echelon Form (REF)

Our first goal is to create a triangular form. We'll start by eliminating the first element in the last row:- Subtract row 1 from row 3: \[ R_3 = R_3 - 2R_1 \]The matrix becomes:\[ \begin{bmatrix} 1 & 2 & -1 & | & -2 \ 1 & 0 & 1 & | & 0 \ 0 & -5 & 1 & | & 1 \end{bmatrix} \]Next, we'll zero out the element in row 2, column 1:- Subtract row 1 from row 2: \[ R_2 = R_2 - R_1 \]The resulting matrix is:\[ \begin{bmatrix} 1 & 2 & -1 & | & -2 \ 0 & -2 & 2 & | & 2 \ 0 & -5 & 1 & | & 1 \end{bmatrix} \]

Step 3: Continue Row Operations to Achieve Reduced Row Echelon Form (RREF)

To simplify further, eliminate the leading coefficient in row 3, column 2:- Add 2.5 times row 2 to row 3: \[ R_3 = R_3 + 2.5R_2 \]The matrix now is:\[ \begin{bmatrix} 1 & 2 & -1 & | & -2 \ 0 & -2 & 2 & | & 2 \ 0 & 0 & 6 & | & 6 \end{bmatrix} \]Next, simplify row 2 by dividing by -2:- \[ R_2 = \frac{1}{-2}R_2 \]This gives:\[ \begin{bmatrix} 1 & 2 & -1 & | & -2 \ 0 & 1 & -1 & | & -1 \ 0 & 0 & 6 & | & 6 \end{bmatrix} \]

Step 4: Solve Backward to Find the Solution

Now that we have a triangular form, we can solve for the variables:- From row 3, simplify the equation: \[ 6z = 6 \rightarrow z = 1 \]- Substitute \( z = 1 \) back into row 2: \[ y - 1 = -1 \rightarrow y = 0 \]- Substitute \( y = 0 \) and \( z = 1 \) back into row 1: \[ x + 2(0) - 1 = -2 \rightarrow x = -1 \]The solution is: \( x = -1 \), \( y = 0 \), and \( z = 1 \).

Step 5: Verify the Solution

Substituting \( x = -1 \), \( y = 0 \), and \( z = 1 \) back into the original equations:1. For \( x + 2y - z = -2 \): \[ -1 + 2(0) - 1 = -2 \]2. For \( x + z = 0 \): \[ -1 + 1 = 0 \]3. For \( 2x - y - z = -3 \): \[ 2(-1) - 0 - 1 = -3 \]Each equation is satisfied, confirming the solution is correct.

Key Concepts

Unique Solution

A system of linear equations can have either one unique solution, no solution, or infinitely many solutions. When a system has a unique solution, it means there is exactly one set of values for the variables that satisfies all the equations simultaneously.
In the problem given, we see a system of three equations with three variables \(x\), \(y\), and \(z\). Our goal using Gaussian elimination is to manipulate these equations into a form where we can identify easily the single solution. If we can use row operations to transform the system into reduced row echelon form (RREF), and each variable leads to a specific value, then the system is said to have a unique solution.
Understanding whether a system has a unique solution depends on its matrix form. Specifically, if the matrix of coefficients is invertible or if the determinant is non-zero, it generally means a unique solution exists.

Augmented Matrix

When dealing with systems of equations, we often use matrices to simplify our calculations. An augmented matrix is an efficient way to represent a system of linear equations. It combines the matrix of coefficients with the constants from the equations' right-hand side into a single matrix. This helps us to perform transformations easily without rewriting the entire system.
For the system provided, the augmented matrix looks like this:

• The first three columns represent the coefficients of \(x\), \(y\), and \(z\) in the equations.
• The last column, after the line, stands for the constants on the right side of the equations.

The augmented matrix for our system gives us a compact way to apply row operations and work towards a solution. It consolidates the problem into a manageable format, paving the path for methods like Gaussian elimination and Gauss-Jordan elimination.

Row Operations

The key to solving systems of equations through Gaussian elimination is the application of row operations. These are transformations applied to rows of the augmented matrix to simplify it while ensuring the system of equations remains consistent. Row operations are essential tools as they facilitate the transition from an augmented matrix to its reduced row echelon form.
There are three primary types of row operations:

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

These operations help in creating zeros below pivot positions, gradually leading the matrix towards a form where the solution is apparent. For example, making zeros below a leading ones (diagonal of the matrix) is crucial for solving via back-substitution as seen at the end of Gaussian elimination.

Reduced Row Echelon Form

Reduced Row Echelon Form (RREF) is the stage of row reduction where each pivot is 1, and all elements in the pivot column, except for the pivot itself, are zeros. Achieving RREF means the system of equations is fully simplified, making it easy to extract the solution.
Some characteristics of RREF include:

• Each leading entry in a row is 1.
• Leading 1's appear to the right of any leading 1's in the rows above.
• Each leading 1 is the only non-zero entry in its column.
• Any row containing only zeros appears at the bottom of the matrix.

In the provided exercise, reaching RREF allowed us to clearly write the equations of the system in such a form that gave the solution directly: \(x = -1\), \(y = 0\), and \(z = 1\). This clear representation makes understanding and solving linear systems efficient and straightforward.