Problem 31 - Algebra and Trigonometry

Question

\(17-36\) . Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x+y-z &=0 \\ x+2 y-3 z &=-3 \\ 2 x+3 y-4 z &=-3 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system is consistent with infinitely many solutions: \( x = 3 - z, y = 2z - 3, z = z \).

Step 1: Setup the Augmented Matrix

First, we convert the given system of linear equations into an augmented matrix. The system of equations is: \(\begin{align*} x+y-z &=0 \ x+2y-3z &=-3 \ 2x+3y-4z &=-3 \end{align*}\) This corresponds to the augmented matrix: \[ \begin{bmatrix} 1 & 1 & -1 & | & 0 \ 1 & 2 & -3 & | & -3 \ 2 & 3 & -4 & | & -3 \end{bmatrix} \]

Step 2: Perform Row Operations

Here, we aim to use row operations to transform the matrix to Row Echelon Form (REF). First, subtract Row 1 from Row 2 to eliminate the first entry of the second row. Then, subtract twice Row 1 from Row 3. The matrix becomes: \[ \begin{bmatrix} 1 & 1 & -1 & | & 0 \ 0 & 1 & -2 & | & -3 \ 0 & 1 & -2 & | & -3 \end{bmatrix} \]

Step 3: Further Simplify the Matrix

Notice that Row 2 and Row 3 are identical. Subtract Row 2 from Row 3 to make Row 3 all zeros: \[ \begin{bmatrix} 1 & 1 & -1 & | & 0 \ 0 & 1 & -2 & | & -3 \ 0 & 0 & 0 & | & 0 \end{bmatrix} \] This matrix is now in Row Echelon Form.

Step 4: Interpret the Solutions

From the REF matrix, we can interpret the solutions. The third row is \(0 = 0\), showing no contradiction, so the system is consistent. We have:- From Row 2: \( y - 2z = -3 \) - From Row 1: \( x + y - z = 0 \)Let's express \(y\) from Row 2: \( y = 2z - 3 \). Substitute \( y = 2z - 3 \) into Row 1:\[ x + (2z - 3) - z = 0 \] \[ x + z - 3 = 0 \] \[ x = 3 - z \]

Step 5: Parametrize the Solution

The solution in terms of parameter \( z \) is:- \( x = 3 - z \)- \( y = 2z - 3 \)- \( z = z \)This indicates that the system has infinitely many solutions with \( z \) as a parameter.

Key Concepts

Augmented Matrix

When solving a system of linear equations, one effective approach is to use the concept of an augmented matrix. This matrix is essentially a compact way to write down the system of equations. Each row of the matrix corresponds to one equation in the system.
The coefficients of each variable in the equation fill the columns of the matrix. An additional column is used for the constants on the right side of the equations.
For instance, consider the system:

• Equation 1: \( x + y - z = 0 \)
• Equation 2: \( x + 2y - 3z = -3 \)
• Equation 3: \( 2x + 3y - 4z = -3 \)

This simplifies into the augmented matrix form:\[\begin{bmatrix}1 & 1 & -1 & | & 0 \1 & 2 & -3 & | & -3 \2 & 3 & -4 & | & -3\end{bmatrix}\]This matrix is helpful because it allows easy visualization and manipulation to solve the equations.

Row Operations

Row operations are essential in systematically solving the equations represented in an augmented matrix. They include adding, subtracting, or swapping rows, and multiplying a row by a non-zero scalar. These operations are applied to move towards a simpler matrix form.
The primary goal of using row operations is to reach

• Row Echelon Form (REF)
• Further, if needed, to reach Reduced Row Echelon Form (RREF)

In this exercise, performing row operations allowed us to simplify this matrix:
Starting with:\[\begin{bmatrix}1 & 1 & -1 & | & 0 \1 & 2 & -3 & | & -3 \2 & 3 & -4 & | & -3\end{bmatrix}\]By subtracting Row 1 from Row 2, and twice Row 1 from Row 3, we modify it:\[\begin{bmatrix}1 & 1 & -1 & | & 0 \0 & 1 & -2 & | & -3 \0 & 1 & -2 & | & -3\end{bmatrix}\]Further simplifying by subtracting Row 2 from Row 3 results in another expression:\[\begin{bmatrix}1 & 1 & -1 & | & 0 \0 & 1 & -2 & | & -3 \0 & 0 & 0 & | & 0\end{bmatrix}\]

Parametric Solution

A parametric solution expresses the solution set of a system of equations in terms of free variables. These free variables are the parameters that can take any real value, leading to infinitely many solutions.
From the row-echelon form of the matrix, we extract equations:

• From Row 1: \( x + y - z = 0 \)
• From Row 2: \( y - 2z = -3 \)
• Row 3 offers no new information since it simplifies to 0

Since the system is consistent, we can express \( y \) and \( x \) in terms of \( z \), leading to parametric forms:

• \( y = 2z - 3 \)
• Substituting \( y \) into Row 1 gives: \( x = 3 - z \)
• \( z = z \)

This solution suggests that for any real value of \( z \), there is a corresponding \( x \) and \( y \) that satisfies the original equations.

Row Echelon Form

Row Echelon Form (REF) transforms an augmented matrix into a specific structured format aiding in solution finding. In REF:

• Every non-zero row has more leading zeros than the row above it
• The leading entry of each non-zero row is 1
• All rows with only zeros are at the bottom

The initial processes to achieve this involve narrowing down the number of variables, ultimately simplifying the solving process.
In our example, after applying row operations, the matrix reaches REF:\[\begin{bmatrix}1 & 1 & -1 & | & 0 \0 & 1 & -2 & | & -3 \0 & 0 & 0 & | & 0\end{bmatrix}\]
This aligns with the rules of REF, guiding us toward a clearer solution path. REF is crucial as it provides structure to analyze whether the system is consistent and how to proceed in computing the solution.