Problem 25 - Precalculus Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$\left\\{\begin{aligned} y-2 z &=0 \\ 2 x+3 y &=2 \\ -x-2 y+z &=-1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution to the system is \( x = 1, y = 0, z = 0 \).

Step 1: Write the Augmented Matrix

We start by writing the augmented matrix of the given system. The coefficient matrix is derived from the equations:\[\begin{bmatrix}0 & 1 & -2 & | & 0 \2 & 3 & 0 & | & 2 \-1 & -2 & 1 & | & -1 \\end{bmatrix}\]

Step 2: Perform Row Operations

We need to transform the matrix into row-echelon form using elementary row operations. Let's swap Row 1 and Row 2 to have a leading 1:\[\begin{bmatrix}2 & 3 & 0 & | & 2 \0 & 1 & -2 & | & 0 \-1 & -2 & 1 & | & -1 \\end{bmatrix}\]Next, let's eliminate the first column below the first pivot (2). We do this by replacing Row 3 with \( R_3 + \frac{1}{2} R_1 \):\[\begin{bmatrix}2 & 3 & 0 & | & 2 \0 & 1 & -2 & | & 0 \0 & \frac{1}{2} & 1 & | & 0 \\end{bmatrix}\]

Step 3: Make Row 1 Pivot to 1

Divide Row 1 by 2 to make the leading coefficient (pivot) 1:\[\begin{bmatrix}1 & \frac{3}{2} & 0 & | & 1 \0 & 1 & -2 & | & 0 \0 & \frac{1}{2} & 1 & | & 0 \\end{bmatrix}\]

Step 4: Eliminate Second Column in Third Row

Subtract \(\frac{1}{2} \times \text{Row 2}\) from \text{Row 3} to eliminate the \(y\) term in the third row:\[\begin{bmatrix}1 & \frac{3}{2} & 0 & | & 1 \0 & 1 & -2 & | & 0 \0 & 0 & 2 & | & 0 \\end{bmatrix}\]

Step 5: Solve for Variables

We now have an upper triangular matrix, which allows us to solve for \(z\) first:- From Row 3: \( 2z = 0 \Rightarrow z = 0 \)- Substitute \(z = 0\) in Row 2: \( y - 2(0) = 0 \Rightarrow y = 0 \)- Substitute \(y = 0\) in Row 1: \( x + \frac{3}{2}(0) = 1 \Rightarrow x = 1 \)

Step 6: Verify the Solution

Substitute \(x = 1, y = 0, z = 0\) back into the original equations to verify:1. \( y - 2z = 0 \Rightarrow 0 - 2(0) = 0 \)2. \( 2x + 3y = 2 \Rightarrow 2(1) + 3(0) = 2 \)3. \( -x - 2y + z = -1 \Rightarrow -(1) - 2(0) + 0 = -1 \)The values satisfy all original equations.

Key Concepts

Augmented Matrix

An augmented matrix provides a neat way to represent a system of linear equations. It is like making a table where you put all the number parts (coefficients) from your equations neatly together.
In an augmented matrix, each row of numbers corresponds to one of the original equations. The left side of the vertical line contains the coefficients of the variables, while the right side of the line contains the constants from each equation.

• For example, from the equations given in this exercise, you derive this matrix representation: \[\begin{bmatrix}0 & 1 & -2 & | & 0 \2 & 3 & 0 & | & 2 \-1 & -2 & 1 & | & -1 \end{bmatrix}\]
• The vertical line in the matrix visually separates the coefficients from the constants.

Using an augmented matrix helps to organize your work and makes it easier to perform row operations, which are crucial in solving the system.

Row Operations

Row operations are strategies used to modify an augmented matrix to make solving equations easier. They include methods like swapping rows, multiplying a row by a non-zero number, and adding or subtracting rows.
These operations aim to transform the matrix into a simpler form, usually row-echelon form or reduced row-echelon form, without changing the solution of the system.

• The first move might be to swap rows. This is handy when you want a row to start with a smaller number or even a zero.
• Next, we might multiply a row by a constant. For instance, you might divide a whole row by 2 to make the numbers smaller and easier to work with.
• Finally, adding or subtracting rows can help you eliminate variables in specific columns, simplifying the matrix.

Performing these operations correctly brings clarity, making the last steps of solving for variables easier.

System of Linear Equations

A system of linear equations is a set of two or more equations that all contain the same variables. The aim is to find values for these variables that satisfy all the equations at once.
Solving these systems helps us understand situations where multiple conditions have to be true at the same time. For example, in this exercise, we are given: \[ \left\{\begin{aligned} y-2 z &=0 \ 2 x+3 y &=2 \ -x-2 y+z &=-1 \end{aligned}\right.\]

• The solution process involves expressing this system in matrix form, making insightful row operations, and using back substitution.
• The ultimate goal is to determine whether a unique solution, no solution, or infinitely many solutions exist for the variables.

Inconsistent systems, where no solution can be found, occur when the equations contradict each other completely. Understanding and solving systems of linear equations is fundamental in linear algebra and many real-world applications.