Problem 10 - Precalculus

Question

In Exercises \(9-26\), put each system of linear equations into triangular form and solve the system if poesible. Classify each system as consistent independent, consistent dependent, or inconsistent. $$ \left\\{\begin{aligned} x+y+z &=3 \\ 2 x-y+z &=0 \\ -3 x+5 y+7 z &=7 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system is consistent independent with solution \((x, y, z) = (1, 2, 0)\).

Step 1: Write Down the System in Matrix Form

Start by representing the system of equations using a matrix. The given system can be written in augmented matrix form as follows:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \2 & -1 & 1 & | & 0 \-3 & 5 & 7 & | & 7\end{bmatrix}\]

Step 2: Apply Row Operations to Achieve Triangular Form

Our goal is to make the augmented matrix into upper triangular form using row operations. Begin by eliminating the coefficients below the first pivot (which is 1 in the first row).- Subtract 2 times the first row from the second row to make the first coefficient of the second row equal zero.- Add 3 times the first row to the third row to make the first coefficient of the third row equal zero.\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & -3 & -1 & | & -6 \0 & 8 & 10 & | & 16\end{bmatrix}\]

Step 3: Continue Row Operations for Triangular Form

Next, eliminate the non-zero coefficient below the pivot in the second column. - Add \(\frac{8}{3}\) times the second row to the third row to make the coefficient in the third row's second column zero.\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & -3 & -1 & | & -6 \0 & 0 & \frac{2}{3} & | & \frac{0}{3}\end{bmatrix}\]This transforms our system into an upper triangular form.

Step 4: Write the System from the Triangular Matrix

From the upper triangular matrix, write each equation as follows:1. \(x + y + z = 3\)2. 3. \(-3y - z = -6\)4. \(\frac{2}{3}z = 0\)

Step 5: Solve the Equations from the Triangular Form

Solving the third equation: \(\frac{2}{3}z = 0\) gives \(z = 0\).Plug \(z = 0\) into the second equation: \[-3y - 0 = -6\]This simplifies to \(y = 2\).Finally, substitute \(y = 2\) and \(z = 0\) into the first equation:\[x + 2 + 0 = 3\]This gives \(x = 1\).

Step 6: Classify the System

Since we found a unique solution \((x, y, z) = (1, 2, 0)\), the system is consistent and independent.

Key Concepts

Triangular Form

Triangular form is a structure for a set of linear equations represented in matrix form where the coefficients below each pivot position are zeros. This is particularly useful because it simplifies solving the system using back substitution.

A system of equations is in triangular form when:

• The first non-zero number from the left (called the pivot) in each non-zero row is a 1.
• Each pivot is further to the right than the pivot in the row directly above it.
• All coefficients below each pivot are zeros.

This form is essential as it directly leads to solving systems of equations by making back substitution a straightforward task. In the example problem, we simplify the system of equations into a triangular form, which makes solving for the variables clear and structured.

Matrix Representation

Matrix representation involves expressing a system of linear equations as a matrix, known as an augmented matrix. It includes coefficients of the variables and constants from each equation in the system.

For example, take the system of equations:

• \( x + y + z = 3 \)
• \( 2x - y + z = 0 \)
• \( -3x + 5y + 7z = 7 \)

The corresponding augmented matrix is:\[\begin{bmatrix} 1 & 1 & 1 & | & 3 \2 & -1 & 1 & | & 0 \-3 & 5 & 7 & | & 7 \end{bmatrix}\]This matrix represents the system concisely and allows the use of row operations to manipulate the system into triangular form. The vertical line separates the coefficients from the constant terms of the equations.

Row Operations

Row operations are used to simplify a matrix to either triangular or reduced row-echelon form, making solving the system straightforward. These operations include:

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

In the solution process:

1. We subtract 2 times the first row from the second row to make the first entry of the second row a zero, and add 3 times the first row to the third row for the same purpose.
2. We then perform the operation of adding \(\frac{8}{3}\) times the second row to the third row to eliminate another coefficient.

The essence of these operations is to strategically simplify the matrix, using the freedom to manipulate each row, thereby achieving the triangular form.

Consistent Independent Systems

A consistent independent system has exactly one solution, meaning that all the equations intersect at a single point in space. From the triangular form, if each equation resolves to a unique value for each variable, the system can be classified as consistent and independent.

In our example, once reduced to triangular form, the system of equations:

• \( x + y + z = 3 \)
  \( -3y - z = -6 \)
  \( \frac{2}{3}z = 0 \)

results in the unique solution \((x, y, z) = (1, 2, 0)\).

This indicates that the original system is consistent and independent, confirming that these equations don't overlap more than at this specific point or diverge without any intersection. Such systems are robustly solvable using matrix methods.