Problem 28 - Algebra and Trigonometry

Question

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: \( x = 1 \), \( y = -1 \), \( z = 0 \).

Step 1: Write the Augmented Matrix

First, we'll write the system of equations as an augmented matrix. The given system is:\[ \begin{cases} x - 2y + z = 3 \ 2x - 5y + 6z = 7 \ 2x - 3y + 2z = 5 \end{cases} \] The augmented matrix becomes: \[ \begin{bmatrix} 1 & -2 & 1 & | & 3 \ 2 & -5 & 6 & | & 7 \ 2 & -3 & 2 & | & 5 \end{bmatrix} \]

Step 2: Row Echelon Form

Our goal is to use elementary row operations to convert the matrix to row echelon form. First, we'll subtract twice the first row from the second row: \( R_2 = R_2 - 2R_1 \) resulting in: \[ \begin{bmatrix} 1 & -2 & 1 & | & 3 \ 0 & -1 & 4 & | & 1 \ 2 & -3 & 2 & | & 5 \end{bmatrix} \] Next, subtract twice the first row from the third row: \( R_3 = R_3 - 2R_1 \): \[ \begin{bmatrix} 1 & -2 & 1 & | & 3 \ 0 & -1 & 4 & | & 1 \ 0 & 1 & 0 & | & -1 \end{bmatrix} \]

Step 3: Simplify the Matrix

Now, we perform row operations to simplify further. We add the second row to the third row: \( R_3 = R_3 + R_2 \): \[ \begin{bmatrix} 1 & -2 & 1 & | & 3 \ 0 & -1 & 4 & | & 1 \ 0 & 0 & 4 & | & 0 \end{bmatrix} \] Simplify the third row by dividing by 4: \( R_3 = \frac{1}{4}R_3 \): \[ \begin{bmatrix} 1 & -2 & 1 & | & 3 \ 0 & -1 & 4 & | & 1 \ 0 & 0 & 1 & | & 0 \end{bmatrix} \]

Step 4: Back Substitution

Now that the matrix is in row echelon form, we can use back substitution to solve for the variables. From the third row: \( z = 0 \). Substitute \( z = 0 \) in the second row equation: \[ -y + 4(0) = 1 \] which simplifies to \( y = -1 \). Substitute \( y = -1 \) and \( z = 0 \) in the first row equation: \[ x - 2(-1) + 0 = 3 \] Simplifying gives \( x = 1 \).

Step 5: Conclusion

The solutions are \( x = 1 \), \( y = -1 \), and \( z = 0 \). Therefore, the system is consistent and has a unique solution.

Key Concepts

Augmented Matrix

An augmented matrix is a helpful way of organizing a linear system's coefficients and constants. It allows us to use matrix operations to simplify solving the system. In our exercise, the original system of equations can be expressed as an augmented matrix:

• The coefficients of the variables create the left part of the matrix.
• The constants from the equations form the right side. This combination is separated by a vertical line.
• For example, the system: \[ \begin{cases} x - 2y + z = 3 \ 2x - 5y + 6z = 7 \ 2x - 3y + 2z = 5 \end{cases} \] becomes the augmented matrix: \[ \begin{bmatrix} 1 & -2 & 1 & | & 3 \ 2 & -5 & 6 & | & 7 \ 2 & -3 & 2 & | & 5 \end{bmatrix} \]

This form simplifies the representation of equations and prepares us for applying row operations next.

Row Echelon Form

The row echelon form is a specific format for matrices that helps in solving linear systems. It's achieved through a series of row operations to simplify the matrix. The goal is to create a staircase-like pattern of zeros below the pivots (leading coefficients) in each row, making back substitution possible. Here's how we transform a matrix into row echelon form:

• Elementary Row Operations: These include swapping rows, multiplying a row by a non-zero constant, or adding/subtracting the multiples of one row from another.
• In our example, after performing operations on our augmented matrix, we ended up with: \[ \begin{bmatrix} 1 & -2 & 1 & | & 3 \ 0 & -1 & 4 & | & 1 \ 0 & 0 & 1 & | & 0 \end{bmatrix} \]
• Note how each pivot (first non-zero number from the left in a row) has only zeros below it.

Reaching this step substantially simplifies the next process of solving for the variables.

Back Substitution

Once a matrix has been transformed into row echelon form, back substitution is used to find the solutions for the variables in the system.

• Start solving for the variable in the last row. In our example, the equation from the last row was \( z = 0 \).
• After finding \( z \), substitute it into the second row equation to solve for \( y \): \( -y + 4(0) = 1 \) simplifies to \( y = -1 \).
• Finally, substitute both \( y = -1 \) and \( z = 0 \) into the first row equation to solve for \( x \): \( x - 2(-1) + 0 = 3 \), leading to \( x = 1 \).

This method traces back through the matrix, using the already determined values to find the remaining unknowns.