Problem 26 - Calculus for Biology and Medicine

Question

In Problems 25-28, find the augmented matrix and use it to solve the system of linear equations. $$ \begin{aligned} -x-2 y+3 z &=-9 \\ 2 x+y-z &=5 \\ 4 x-3 y+5 z &=-9 \end{aligned} $$

Step-by-Step Solution

Answer

\(x = 1/7\), \(y = 10/7\), \(z = 20/7\).

Step 1: Write the System of Equations

The given system of linear equations is:\(-x - 2y + 3z = -9\),\(2x + y - z = 5\),\(4x - 3y + 5z = -9\).

Step 2: Form the Augmented Matrix

Convert the system of equations into an augmented matrix. The coefficients of the variables form the matrix's main part, and the constants form the augmented part:\[\begin{bmatrix}-1 & -2 & 3 & | & -9 \2 & 1 & -1 & | & 5 \4 & -3 & 5 & | & -9\end{bmatrix}\]

Step 3: Apply Row Operations to Obtain Row Echelon Form

Use row operations to simplify the augmented matrix to row-echelon form:1. Swap Row 1 and Row 2:\[\begin{bmatrix}2 & 1 & -1 & | & 5 \-1 & -2 & 3 & | & -9 \4 & -3 & 5 & | & -9\end{bmatrix}\]2. Replace Row 3 with Row 3 - 2 * Row 1:\[\begin{bmatrix}2 & 1 & -1 & | & 5 \-1 & -2 & 3 & | & -9 \0 & -5 & 7 & | & -19\end{bmatrix}\]3. Replace Row 2 with Row 2 + 0.5 * Row 1:\[\begin{bmatrix}2 & 1 & -1 & | & 5 \0 & -1.5 & 2.5 & | & -6.5 \0 & -5 & 7 & | & -19\end{bmatrix}\]4. Replace Row 3 with Row 3 - (10/3) * Row 2:\[\begin{bmatrix}2 & 1 & -1 & | & 5 \0 & -1.5 & 2.5 & | & -6.5 \0 & 0 & -7/3 & | & -20/3\end{bmatrix}\]

Step 4: Solve for Variables

Solve the system from the row echelon form:1. From the 3rd row: \(-\frac{7}{3}z = -\frac{20}{3}\) gives \(z = 20/7\).2. Substitute \(z = 20/7\) in the 2nd row: \(\frac{5}{2}z - \frac{3}{2}y = 13/2\) gives \(y = 10/7\).3. Substitute \(y = 10/7\) and \(z = 20/7\) in the 1st row: \(2x + y - z = 5\) gives \(x = 1/7\).

Step 5: Write the Solution

The solution to the system of equations is the set of values for \(x\), \(y\), and \(z\). Thus, the final solution is:\(x = 1/7\), \(y = 10/7\), \(z = 20/7\).

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations with the same set of variables. For our exercise, the given system includes three equations with the variables, \(x\), \(y\), and \(z\). Each equation can be thought of as a plane in a three-dimensional space. An intersection of these planes gives us the solution of the system, which is the point or set of points that satisfy all equations simultaneously.

• The equations are typically written in the form: \( ax + by + cz = d \), where \(a, b, c,\) and \(d\) are constants.
• Solving a system means finding the values of \(x, y,\) and \(z\) that make all the equations true at the same time.
• Solutions can be a unique point, infinitely many points (if the planes coincide or intersect along a line), or no solution (if the planes do not intersect at all).

In this particular system, we have ensured that all equations are balanced and aligned correctly so they can be represented in a matrix form, which aids in finding solutions through systematic approaches.

Row Operations

Row operations are procedures used to manipulate the rows of a matrix to achieve a particular form, typically making the matrix easier to work with when solving linear equations. There are three types of row operations you can use:

• Swapping two rows, which allows you to move equations to simplify calculations
• Multiplying a row by a nonzero constant, which enables changing the form of the equation to make coefficients easier to handle
• Adding or subtracting another row from a row, a vital step in combining equations to eliminate variables

In our matrix transformation process, we used these operations several times: 1. Swapping to place equations with easier-to-handle coefficients on top 2. Reducing row values systematically for zero entries below leading coefficients 3. Ultimately, transforming our matrix leads to more straightforward back-solving for the variable's values. These operations are carried out systematically and help us to reach a matrix form where solving for the variables is direct and clear.

Row Echelon Form

The row echelon form of a matrix is a simplified version making it easier to extract solutions. A matrix is in row echelon form when:

• All non-zero rows are above any rows of zeros
• The leading coefficient of a non-zero row is always to the right of the leading coefficient of the row above it
• All elements below a leading coefficient are zeros

This form is crucial because it allows systematic back-substitution to solve for each variable starting from the bottom up. In transforming to row echelon form, you essentially make the system resemble a triangular structure: 1. We encounter a "staircase" pattern of leading variables like steps down from left to right 2. Solving with back-substitution refers to finding the value of the last variable first and then substituting back to find others. Ultimately, this arrangement simplifies handling complex systems and ensures a clear path to the solution, as each step in the echelon form takes us closer to identifying the precise values of the variables.