Problem 22 - Precalculus: Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. \(\left\\{\begin{aligned} x+y+z &=0 \\\\-x+2 y+5 z &=3 \\ 3 x-y &=6 \end{aligned}\right.\)

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (1, -3, 2)\).

Step 1: Write the system as augmented matrix

First, we rewrite the given system of equations as an augmented matrix. The coefficients and constants are: \[\begin{bmatrix} 1 & 1 & 1 & | & 0 \ -1 & 2 & 5 & | & 3 \ 3 & -1 & 0 & | & 6 \end{bmatrix}.\]

Step 2: Perform row operations to get in Row Echelon Form

Add the first row to the second row to eliminate \(x\) in the second row:\[R_2 = R_2 + R_1 \to \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 3 & 6 & | & 3 \ 3 & -1 & 0 & | & 6 \end{bmatrix}.\]Next, multiply the first row by 3 and subtract from the third row to eliminate \(x\) in the third row:\[R_3 = R_3 - 3R_1 \to \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 3 & 6 & | & 3 \ 0 & -4 & -3 & | & 6 \end{bmatrix}.\]

Step 3: Further simplify to reach upper triangular form

Multiply the second row by \(\frac{1}{3}\) to normalize:\[R_2 = \frac{1}{3}R_2 \to \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 1 & 2 & | & 1 \ 0 & -4 & -3 & | & 6 \end{bmatrix}.\]Then add 4 times the second row to the third row to eliminate \(y\):\[R_3 = R_3 + 4R_2 \to \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 1 & 2 & | & 1 \ 0 & 0 & 5 & | & 10 \end{bmatrix}.\]

Step 4: Back-substitution to find solutions

From the third row, we have:\[5z = 10 \Rightarrow z = 2.\]Substitute \(z = 2\) into the second row:\[y + 2(2) = 1 \Rightarrow y = 1 - 4 = -3.\]Substitute \(y = -3\) and \(z = 2\) into the first row to find \(x\):\[x + (-3) + 2 = 0 \Rightarrow x = 1.\]

Step 5: Write the complete solution

The solution to the system is \(x = 1\), \(y = -3\), \(z = 2\). Thus, the point \((1, -3, 2)\) is the complete solution to the system.

Key Concepts

Augmented Matrix

In the realm of linear algebra, an augmented matrix is a powerful tool when solving systems of linear equations. Think of it as a compact form of representing the system, where both the coefficients of the variables and the constants from the equations are included in a single matrix layout.

When you see a system like this:

• \(x+y+z = 0\)
• \(-x+2y+5z = 3\)
• \(3x - y = 6\)

It can be transformed into an augmented matrix:\[\begin{bmatrix} 1 & 1 & 1 & | & 0 \-1 & 2 & 5 & | & 3 \3 & -1 & 0 & | & 6 \end{bmatrix}\]Here, the vertical line separates the coefficients from the constants, providing a structured approach for further operations.

This representation is the starting point in many solution strategies, like Gaussian elimination, as it allows us to perform row operations systematically.

Row Operations

Row operations are the heart of transforming an augmented matrix into forms that make solving it easier. There are three elementary row operations:

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

These operations are used because they do not change the solution set of the system.

For example, suppose we need to eliminate the coefficient of \(x\) in subsequent rows, we can use row operations, like:

• Adding the first row to the second to eliminate the \(x\) term.
• Multiplying a row and then subtracting from another to create zeros where needed.
  

Ultimately, these operations help in systematically simplifying the matrix, often aiming for a row echelon form or even further to a reduced row echelon form.

Back-Substitution

Once an augmented matrix is simplified to the point where it's in an upper triangular form, usually following row operations, it's time to apply back-substitution. This process begins with the bottom-most equation in the triangular matrix form and moves upwards.

In practice, if the last row reads \(5z = 10\), we can immediately solve for \(z\) by dividing both sides by 5 to find \(z = 2\).

The solved value of \(z\) is then substituted back into the equations represented by the earlier rows. Working upwards, use the known values to solve for other variables, step by step. For instance, using \(z = 2\) in the second equation might yield \(y = -3\), continuing the process.

Finally, substituting all known variables into the first row allows for solving \(x\), pulling together the complete solution.

Solution of Linear Equations

Finding the solution to a system of linear equations is the culmination of a series of methodical steps, starting from writing the system as an augmented matrix, performing row operations, and concluding with back-substitution.

The example system gives us a unique solution, which means all equations intersect at exactly one point in space if visualized in a three-dimensional context. For our specific system, the solution is:

• \(x = 1\)
• \(y = -3\)
• \(z = 2\)

Thus, the point \((1, -3, 2)\) satisfies all original equations simultaneously.

When all steps are followed correctly, solutions can always be trusted if they exist. This reliable process showcases the power of systematic approaches in linear algebra.