Problem 67 - Precalculus

Question

Solve the system of linear equations using Gaussian elimination with back- substitution. $$\begin{aligned} &2 y+z=3\\\ &4 x-z=-3\\\ &7 x-3 y-3 z=2\\\ &x-y-z=-2 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \((x,y,z) = \left(-\frac{7}{2}, 7, -11\right)\).

Step 1: Write the System in Matrix Form

First, we write the system of equations as an augmented matrix. The system of equations is:\[\begin{align*}2y + z &= 3 \4x - z &= -3 \7x - 3y - 3z &= 2 \x - y - z &= -2\end{align*}\]In augmented matrix form, it is:\[\begin{bmatrix}0 & 2 & 1 & | & 3 \4 & 0 & -1 & | & -3 \7 & -3 & -3 & | & 2 \1 & -1 & -1 & | & -2 \end{bmatrix}\]

Step 2: Perform Gaussian Elimination

We aim to form an upper triangular matrix where the leading coefficients of each row below the first are zeros. 1. Swap the second and first row to introduce a leading "4" for easier operations:\[\begin{bmatrix}4 & 0 & -1 & | & -3 \0 & 2 & 1 & | & 3 \7 & -3 & -3 & | & 2 \1 & -1 & -1 & | & -2 \end{bmatrix}\]2. Make the first column entries below the leading coefficient to zero by: * Row 3: (7R1 - 4R3) * Row 4: (R4 - R1)Updated matrix:\[\begin{bmatrix}4 & 0 & -1 & | & -3 \0 & 2 & 1 & | & 3 \0 & -3 & -1 & | & 23 \0 & -1 & -1 & | & 1 \end{bmatrix}\]

Step 3: Continue Gaussian Elimination for Upper Triangle

Make the leading coefficient of the second column to zero from below:1. For Row 3: Use R3 - (3/2)R22. For Row 4: Use R4 + (1/2)R2Updated matrix:\[\begin{bmatrix}4 & 0 & -1 & | & -3 \0 & 2 & 1 & | & 3 \0 & 0 & -\frac{5}{2} & | & \frac{33}{2} \0 & 0 & -\frac{1}{2} & | & \frac{11}{2} \end{bmatrix}\]

Step 4: Perform Back-Substitution

Begin solving from the last row up, replacing known values into previously solved equations:1. From Row 4: \(-\frac{1}{2}z = \frac{11}{2}\) implies \(z = -11\)2. Substitute \(z\) in Row 3: \(-\frac{5}{2}z = \frac{33}{2}\) results in a redundant equation; thus no additional information is needed.3. From Row 2: \(2y + z = 3\) substitute \(z = -11\) gives \(2y - 11 = 3\), so \(y = 7\).4. Substitute \(y\) and \(z\) in Row 1: \(4x - z = -3\) results in \(4x + 11 = -3\), giving \(x = -\frac{7}{2}\).

Step 5: Verify the Solution

Verify \((x, y, z) = \left(-\frac{7}{2}, 7, -11\right)\) satisfies each original equation:1. Row 2: \(4(-\frac{7}{2}) - (-11) = -3\) checks.2. Row 1: \(2y + z = 3\) gives \(14 - 11 = 3\) checks.3. Row 3: Validates with substituted values.4. Row 4: \(x - y - z = -2\) gives \(-\frac{7}{2} - 7 + 11 = -2\).Each equation is satisfied with the provided solutions.

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more equations with multiple variables. These equations are linear, which means they graph as straight lines. Solving such a system involves finding values for the variables that satisfy all given equations at once. For example, if you have equations involving variables \(x\), \(y\), and \(z\), solving the system will give you specific values for these variables that make each equation hold true.

To understand whether a solution exists, and to find the solution, one uses methods like substitution, elimination, or matrix operations such as Gaussian elimination. These methods systematically reduce the equations to find the values of the variables. Solving systems of linear equations is foundational to subjects such as algebra and calculus, and it is widely applied in fields like engineering, physics, and economics.

In our original problem, the equations were structured to find the values of \(x\), \(y\), and \(z\) that work for four equations simultaneously. This involved transforming the equations into an augmented matrix and then employing Gaussian elimination.

Back-Substitution

Back-substitution is a key step in solving systems of equations, particularly when using Gaussian elimination. Once you've transformed your system of equations into an upper triangular matrix, you proceed with back-substitution to find the values of the unknowns. The upper triangular matrix allows you to start solving from the last equation, which typically involves only one unknown.

Here’s how it works:

• Start from the last row of your matrix where typically one variable is left, solve for this variable.
• Substitute this value into the above row equations, reducing the number of unknowns as you move up the matrix.
• Continue this process until all variables are solved.

In the original step-by-step solution, back-substitution began after forming the upper triangular matrix. The last row provided the value for \(z\), and this was used in the rows above to sequentially find \(y\) and then \(x\). This reversed substitution process efficiently resolved the variables.

Augmented Matrix

An augmented matrix is a powerful tool used in solving systems of linear equations. It is essentially a compact way to represent a system, combining both the coefficients of the variables and the constants from the right side of the equations. This simplifies operations and makes it easier to apply methods like Gaussian elimination.

To form an augmented matrix:

• List the coefficients of each variable in the equations, aligning them in columns.
• Draw a vertical line to separate these coefficients from the constants on the right side of the equations.

The transformation of systems of equations into an augmented matrix is the first step in methods like Gaussian elimination. In our original exercise, the given equations were arranged into an augmented matrix, which helped in performing row operations systematically. This setup paved the way for simplifying the system into a form where back-substitution could be easily applied to find the solutions.