Problem 35 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understanding the System

We have a system of 4 linear equations with 4 variables \(x\), \(y\), \(z\), and \(w\). We aim to solve this system using either substitution or elimination to find the values of these variables or determine if the system is inconsistent.

Step 2: Equation as Matrix Form

Write the augmented matrix for the system: \[\begin{bmatrix}1 & 0 & 1 & 2 & | 6 \0 & 1 & -2 & 0 & | -3 \1 & 2 & -1 & 0 & | -2 \2 & 1 & 3 & -2 & | 0\end{bmatrix}\]Next, we will use Gaussian elimination to simplify the matrix.

Step 3: Eliminate First Column Below Pivot

Subtract the first row from the third row and 2 times the first row from the fourth row to eliminate the \(x\) variable from these rows:New third row: \([0, 2, -2, -2 | -8]\)New fourth row: \([0, 1, 1, -6 | -12]\)

Step 4: Eliminate Second Column Below Pivot

Subtract twice the second row from the third row to eliminate the \(y\) variable:New third row: \([0, 0, 2, -2 | -2]\).Subtract the second row from the fourth row:New fourth row: \([0, 0, 3, -6 | -9]\)

Step 5: Eliminate Third Column Below Pivot

Scale the third row by dividing everything by 2: \([0, 0, 1, -1 | -1]\)Subtract 3 times the third row from the fourth row to eliminate \(z\):New fourth row: \([0, 0, 0, -3 | -6]\)

Step 6: Back Substitution to Solve

Simplify the fourth row so that it is in the form of \(-3w = -6\), solving gives \(w = 2\). Substitute \(w = 2\) into the third row and solve for \(z\):\(z - 1(2) = -1 \Rightarrow z = 1\).Substitute values of \(z\) and \(w\) into the second row and solve for \(y\):\(y - 2(1) = -3 \Rightarrow y = -1\).Substitute all known values into the first equation to solve for \(x\):\(x + 1 + 2(2) = 6 \Rightarrow x = 1\).

Step 7: Check Consistency

With \(x = 1\), \(y = -1\), \(z = 1\), and \(w = 2\), substitute these values into all original equations to verify that they are satisfied. If the system is consistent, the solution will hold true for all equations.

Key Concepts

Gaussian Elimination

Gaussian elimination is a systematic method used to solve systems of linear equations. It transforms the system into a simpler form, typically an upper triangular matrix, where the coefficients below the main diagonal are zero. This process involves a sequence of operations:

• Swapping rows: To have a non-zero element as a pivot when necessary.
• Multiplying a row by a non-zero scalar: To make pivot values equal to 1 or simplify calculations.
• Adding or subtracting a multiple of one row from another: To eliminate variables below the pivot position.

By applying these operations consistently, you gradually form a matrix that makes the system easier to solve through back substitution.
Gaussian elimination can identify whether a system is consistent (a solution exists) or inconsistent (no solutions exist). For example, if Gaussian elimination results in a row where all the variables vanish but the constant is non-zero, the system is inconsistent.

Back Substitution

Once Gaussian elimination has simplified the system into an upper triangular form, back substitution starts. This part of the process involves solving the equations from the bottom up. You start with the last row which usually contains one variable.
For example, the equation \(-3w = -6\) can be solved directly for \(w\), giving \(w = 2\).
With \(w\) found, substitute it into the preceding rows. Next, solve for \(z\), in \(z - w = -1\), which becomes \(z - 2 = -1\), and thus \(z = 1\).

• Substitute known variables back into other rows.
• Continue upward, solving for each variable.

For the system to be consistent, the values need to satisfy all original equations when substituted back.

Augmented Matrix

An augmented matrix is a helpful tool for simplifying the representation of a system of linear equations. It combines the coefficient matrix with the constants from the equations into a single matrix. The vertical bar in an augmented matrix separates the coefficients of the variables from the constants.
This visual and compact structure allows easier manipulation during Gaussian elimination. For example, in our system: \[\begin{bmatrix}1 & 0 & 1 & 2 & | 6 \0 & 1 & -2 & 0 & | -3 \1 & 2 & -1 & 0 & | -2 \2 & 1 & 3 & -2 & | 0\end{bmatrix}\] Each row corresponds directly to one of the equations in the system.

• It gives you a clearer overview of the whole system.
• Facilitates the use of row operations during Gaussian elimination.

This form is particularly powerful for visualizing and executing the steps needed for both Gaussian elimination and back substitution.

Variable Elimination

Variable elimination removes variables from equations to simplify solving systems of equations. The concept is the cornerstone of Gaussian elimination, where you aim to focus each operation on eliminating one variable at a time.
The process involves strategically applying the row operations to make the coefficients of the variables below the pivot position zero.

• Simplifying equations so fewer variables are considered in each step.
• Making subsequent steps, like back substitution, straightforward by reducing complexity.

Consider, for example, eliminating \(y\) from the system by row operations. You reduce its impact in one or more rows, which makes solving easier. As you progress, you focus on one variable per step, progressively breaking down complex interactions between variables.