Problem 34 - Calculus for Biology and Medicine

Question

In Problems 29-34, determine whether each system is overdetermined or underdetermined; then solve each system. $$ \begin{array}{rr} x+y= & -1 \\ 2 x-y= & 7 \\ x-2 y= & 8 \end{array} $$

Step-by-Step Solution

Answer

The system is overdetermined, and the solution is \(x = 2\), \(y = -3\).

Step 1: Identify the Type of System

An overdetermined system has more equations than unknowns, while an underdetermined system has fewer. Here, we have three equations involving two unknowns, \(x\) and \(y\). Therefore, this system is overdetermined.

Step 2: Set Up the Augmented Matrix

Write the system as an augmented matrix:\[\begin{bmatrix}1 & 1 & | & -1 \2 & -1 & | & 7 \1 & -2 & | & 8\end{bmatrix}\]

Step 3: Perform Row Operations

Transform the matrix using row operations to simplify it as much as possible. First, subtract Row 1 from Row 3:\[R3 = R3 - R1 \Rightarrow \begin{bmatrix}1 & 1 & | & -1 \2 & -1 & | & 7 \0 & -3 & | & 9\end{bmatrix}\]Next, use Row 1 to simplify Row 2:\[R2 = R2 - 2R1 \Rightarrow \begin{bmatrix}1 & 1 & | & -1 \0 & -3 & | & 9 \0 & -3 & | & 9\end{bmatrix}\]

Step 4: Analyze the Matrix

The last two rows are identical, indicating a possibility of dependent equations. We are left with two independent equations from these operations. Essentially, this reduces our system back to:\[\begin{align*}x + y &= -1 \-y &= 3 \end{align*}\]

Step 5: Solve for One Variable

From the second equation, solve for \(y\):\[-y = 3 \Rightarrow y = -3\]

Step 6: Substitute Back to Find the Other Variable

Substitute \(y = -3\) into the first simplified equation:\[x - 3 = -1 \Rightarrow x = 2\]

Step 7: Verify the Solution

Substitute \(x = 2\) and \(y = -3\) back into all original equations to ensure they are satisfied:\[\begin{align*}2 - 3 &= -1 \4 + 3 &= 7 \2 + 6 &= 8\end{align*}\]All equations hold true, verifying our solution.

Key Concepts

Understanding the Augmented Matrix

An augmented matrix is a powerful way to represent a system of linear equations. It involves writing the coefficients of the variables in an organized rectangular array, with a line separating the coefficients from the constants of the equations. For the given problem, we express the system of equations as an augmented matrix: \[ \begin{bmatrix} 1 & 1 & | & -1 \ 2 & -1 & | & 7 \ 1 & -2 & | & 8 \end{bmatrix} \] Here is what each part of the matrix tells us:

• The numbers before the separator (|) are the coefficients for the variables \(x\) and \(y\) in the equations.
• The numbers after the separator represent the constants on the right side of the equation.
• Each row corresponds to an equation from the system.

Using the augmented matrix helps us manage the equations in a structured way, making it easier to apply operations to find the solution.

Simplifying with Row Operations

Row operations are essential tools used to simplify an augmented matrix and find solutions to a system of equations. They allow us to manipulate the matrix while keeping the system's solutions the same. Here’s how we use them:

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

Let’s see these operations in action on our matrix: First, we perform the operation \(R3 = R3 - R1\) to get: \[ \begin{bmatrix} 1 & 1 & | & -1 \ 2 & -1 & | & 7 \ 0 & -3 & | & 9 \end{bmatrix} \] Then, simplify further by \(R2 = R2 - 2R1\): \[ \begin{bmatrix} 1 & 1 & | & -1 \ 0 & -3 & | & 9 \ 0 & -3 & | & 9 \end{bmatrix} \] This process of simplifying the matrix through row operations helps in identifying dependent equations and finding solutions more efficiently.

Exploring Dependent Equations

Dependent equations occur when one equation in a system can be derived from another, meaning they essentially contain the same information. In the matrix we've simplified, notice the resemblance of the second and third rows: \[ \begin{bmatrix} 1 & 1 & | & -1 \ 0 & -3 & | & 9 \ 0 & -3 & | & 9 \end{bmatrix} \] This indicates that these two rows are dependent on each other. Realizing this dependency is crucial because:

• Dependent equations do not increase the dimension of the solution space; they merely confirm solutions provided by other independent equations.
• By identifying dependent equations, we simplify the system, reducing it to independent ones, which are sufficient to determine the solutions.

In our example, simplifying the system to two independent equations, \(x + y = -1\) and \(-y = 3\), allows us to straightforwardly solve for \(x\) and \(y\), verifying the solution for consistency against the original system.