Problem 37 - Contemporary Precalculus

Question

Solve the system by any method. $$\begin{aligned} x+2 y+3 z &=1 \\ 3 x+2 y+4 z &=-1 \\ 2 x+6 y+8 z+w &=3 \\ 2 x+\quad 2 z-2 w &=3 \end{aligned}$$

Step-by-Step Solution

Answer

$$ \left[ \begin{array}{cccc|c} 1 & 2 & 3 & 0 & 1 \\ 0 & 1 & 5/4 & 0 & 1 \\ 0 & 0 & 1/2 & 1 & 2 \\ 0 & 0 & 1/4 & -2 & 2 \\ \end{array} \right] $$ #tag_title# Step 3: Use back-substitution #tag_content# Now that we have the triangular form of the matrix, we will use back-substitution to find the values of the variables. Begin by solving for z from the third row: - z = 4*(1/2) - 2 = 0 Next, solve for y using the value of z in the second row: - y + 5/4 * z = 1 --> y = 1 Finally, use the values of y and z to find x from the first row: - x + 2*y + 3*z = 1 --> x = 1 - 2*1 - 3*0 = -1 The solution of the given system of linear equations is x = -1, y = 1, and z = 0. #ShortAnswer# The solution of the given system of linear equations is x = -1, y = 1, and z = 0.

Step 1: Form the augmented matrix

Write the augmented matrix corresponding to the given system of linear equations. Each row of the matrix represents one equation, with the coefficient of each variable making up the first four columns, and the constant terms in the fifth column. $$ \left[ \begin{array}{cccc|c} 1 & 2 & 3 & 0 & 1 \\ 3 & 2 & 4 & 0 & -1 \\ 2 & 6 & 8 & 1 & 3 \\ 2 & 0 & 2 & -2 & 3 \\ \end{array} \right] $$

Step 2: Perform row operations

Carry out row operations to rewrite the augmented matrix into a triangular form. First, eliminate the x-coefficient in the second, third, and fourth rows: - R2 = R2 - 3 * R1 - R3 = R3 - 2 * R1 - R4 = R4 - 2 * R1 $$ \left[ \begin{array}{cccc|c} 1 & 2 & 3 & 0 & 1 \\ 0 & -4 & -5 & 0 & -4 \\ 0 & 2 & 2 & 1 & 1 \\ 0 & -4 & -4 & -2 & 1 \\ \end{array} \right] $$ Now, eliminate the y-coefficient in the third and fourth rows: - R2 = -1/4 * R2 (to make y-coefficient in R2 equal to 1) - R3 = R3 + 1/2 * R2 - R4 = R4 + R2

Key Concepts

Augmented Matrix

To solve a system of linear equations, we often use an augmented matrix. An augmented matrix includes coefficients of all the variables from each equation as well as the constant terms on the right side of the equations. This matrix acts as a compact, organized way to depict a system of equations.
Let's take an example of the system of equations:

• The first row in the augmented matrix corresponds to the equation \( x + 2y + 3z = 1 \), so it appears as \([1, 2, 3, 0 | 1]\).
• The second row corresponds to \( 3x + 2y + 4z = -1 \) appearing as \([3, 2, 4, 0 | -1]\).
• Similarly, the remaining equations are represented by their respective coefficients.

By turning a system of equations into an augmented matrix, we set up a system that's much simpler to work with when performing calculations like row operations.

Row Operations

Once we have the augmented matrix, we use row operations to simplify it and solve the system of equations. Row operations involve modifying the rows of a matrix to make solving for the variables easier. They include:

• Swapping two rows: Changing the order of equations.
• Multiplying a row by a non-zero number: Useful to normalize a pivot position.
• Adding or subtracting a multiple of one row to another: Helps in eliminating a variable from an equation.

In our example, the goal is to modify the augmented matrix such that one of the equations (rows) largely cancels out a variable in the subsequent equations. This process significantly simplifies the system and makes it easier to reach a solution. For instance, we used \( R2 = R2 - 3 \times R1 \) to eliminate the \( x \)-coefficient from the second row, which pushed us closer to solving the system.

Triangular Form

The triangular form of a matrix means that below its main diagonal, all elements are zeros. Solving a system of equations becomes easier when the augmented matrix is converted into this form.
In our solution example, we aim for a triangular form through row operations. This involves rearranging and adjusting the coefficients so that the equations have a step-like pattern.
This pattern allows us to use back substitution efficiently:

• Upper Triangle vs. Lower Triangle: We focus on an upper triangular form where the lower part of the matrix is zero. This aligns with solving for variables from the bottom up.
• Efficient Solving: Once in triangular form, you can solve the last variable directly, substitute it back into the earlier equations, and acquire all variable values sequentially.

Triangularization simplifies many complex systems, showcasing why it's a preferred technique when solving systems of linear equations.