Problem 32 - Algebra and Trigonometry

Question

\(17-36\) . Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x-2 y+z &=3 \\ 2 x-5 y+6 z &=7 \\ 2 x-3 y-2 z &=5 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system has a unique solution: \(x = \frac{11}{8}, y = -\frac{7}{8}, z = -\frac{1}{8}\).

Step 1: Write the system in matrix form

The given system of equations can be expressed as an augmented matrix:\[\begin{bmatrix}1 & -2 & 1 & | & 3 \2 & -5 & 6 & | & 7 \2 & -3 & -2 & | & 5\end{bmatrix}\] This matrix represents the coefficients and constants of each equation.

Step 2: Use row operations to simplify

To simplify the matrix, first use row operations to create zeros below the first pivot position (first column, first row):1. Subtract the first row from the second row to find a new second row.2. Subtract the first row from the third row to find a new third row.This yields:\[\begin{bmatrix}1 & -2 & 1 & | & 3 \0 & -1 & 5 & | & 1 \0 & -1 & -3 & | & 2\end{bmatrix}\]

Step 3: Eliminate below second pivot position

Subtract the new second row from the third row to eliminate the variable in the second column:\[\begin{bmatrix}1 & -2 & 1 & | & 3 \0 & -1 & 5 & | & 1 \0 & 0 & -8 & | & 1\end{bmatrix}\] Now, the matrix is in upper triangular form.

Step 4: Back-substitute to find solutions

Start back substitution from the bottom:From the third row: \(-8z = 1\) So, \(z = -\frac{1}{8}\).Substitute \(z = -\frac{1}{8}\) into the second row:\[0 * y + 5 * \left(-\frac{1}{8}\right) = 1\] \(-y = \frac{7}{8}\) \(y = -\frac{7}{8}\).Substitute \(y = -\frac{7}{8}\) and \(z = -\frac{1}{8}\) into the first row:\[x - 2 * \left(-\frac{7}{8}\right) + \left(-\frac{1}{8}\right) = 3\]\(x + \frac{7}{4} - \frac{1}{8} = 3\) \(x = 3 - \frac{7}{4} + \frac{1}{8}\).Calculate to solve for \(x\).

Step 5: Solve for the exact solutions

Calculate the expression: \[x = 3 - \frac{7}{4} + \frac{1}{8} \]Common denominator is 8:\[x = \frac{24}{8} - \frac{14}{8} + \frac{1}{8} \]Thus, \[x = \frac{11}{8}\]So the solution to the system is:\(x = \frac{11}{8}, y = -\frac{7}{8}, z = -\frac{1}{8}\).

Key Concepts

Augmented Matrix

Transforming a system of linear equations into an augmented matrix is the first step toward solving it using matrix operations. An augmented matrix combines the coefficients of the variables and the constants from the equations into a single matrix. This makes it easier to apply matrix methods to find solutions.

For example, the system:

• Equation 1: \( x - 2y + z = 3 \)
• Equation 2: \( 2x - 5y + 6z = 7 \)
• Equation 3: \( 2x - 3y - 2z = 5 \)

can be represented as the augmented matrix:\[\begin{bmatrix} 1 & -2 & 1 & | & 3 \ 2 & -5 & 6 & | & 7 \ 2 & -3 & -2 & | & 5 \end{bmatrix}\]The vertical line separates the coefficients from the constants. This simplification is crucial for using row operations effectively.

Row Operations

Row operations are fundamental tools to transform matrices while solving linear equations. The goal is to simplify the augmented matrix to a point where the solution can be easily identified. These operations include:

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

In our example, subtracting the first row from the second and third rows simplified the matrix:\[\begin{bmatrix} 1 & -2 & 1 & | & 3 \ 0 & -1 & 5 & | & 1 \ 0 & -1 & -3 & | & 2 \end{bmatrix}\]These steps helped eliminate variables systematically, leading us to a simpler form.

Back Substitution

Back substitution is the process used when a matrix is transformed into an upper triangular form, where all elements below the main diagonal are zeros. This step is crucial as it allows for solving equations starting from the bottom and working upwards.
In the exercise, once the matrix becomes:\[\begin{bmatrix} 1 & -2 & 1 & | & 3 \ 0 & -1 & 5 & | & 1 \ 0 & 0 & -8 & | & 1 \end{bmatrix}\]we then solve for each variable starting from the last equation where \(-8z = 1 \), giving us \( z = -\frac{1}{8} \).Substituting back into the equation above allows us to find \( y \), and further substitution reveals \( x \).
This method ensures all variables are calculated correctly by solving each equation one by one, avoiding errors.

Upper Triangular Matrix

An upper triangular matrix has all zeros below the main diagonal, making it a crucial form for solving linear systems via back substitution. This configuration greatly simplifies the solution process.
The transition to an upper triangular form occurs through strategic use of row operations. Once the matrix:\[\begin{bmatrix} 1 & -2 & 1 & | & 3 \ 0 & -1 & 5 & | & 1 \ 0 & 0 & -8 & | & 1 \end{bmatrix}\]is achieved, solving for each variable becomes straightforward.
This method streamlines complex problems by reducing them to simpler steps, highlighting the power of matrix transformation in algebraic problem-solving.