Problem 28 - Precalculus: Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. \(\left\\{\begin{array}{rr}y-z= & -1 \\ 6 x+2 y+z= & 2 \\ -x-y-3 z= & -2\end{array}\right.\)

Step-by-Step Solution

Answer

The solution is \(x = -\frac{1}{6}\), \(y = \frac{1}{2}\), \(z = \frac{3}{2}\).

Step 1: Write the System as Augmented Matrix

Start by writing the given equations as an augmented matrix: \[\begin{bmatrix}0 & 1 & -1 & | & -1 \6 & 2 & 1 & | & 2 \-1 & -1 & -3 & | & -2\end{bmatrix}\].

Step 2: Row Reduce to Echelon Form

Use row operations to transform the matrix to its row-echelon form. First, swap row 1 and row 2:\[\begin{bmatrix}6 & 2 & 1 & | & 2 \0 & 1 & -1 & | & -1 \-1 & -1 & -3 & | & -2\end{bmatrix}\]Next, eliminate the leading coefficient below the first entry of the current first row by adding row 3 to row 1 six times:\[\begin{bmatrix}6 & 2 & 1 & | & 2 \0 & 1 & -1 & | & -1 \0 & -2 & -2 & | & -4\end{bmatrix}\].

Step 3: Simplify Row 3

To eliminate the y-term from row 3, add twice row 2 to row 3:\[\begin{bmatrix}6 & 2 & 1 & | & 2 \0 & 1 & -1 & | & -1 \0 & 0 & -4 & | & -6\end{bmatrix}\].

Step 4: Solve for Variables Using Back Substitution

The system of equations that corresponds to this matrix is:1. \(6x + 2y + z = 2\)2. \(y - z = -1\)3. \(-4z = -6\)Solve equation 3 for \(z\): \[z = \frac{3}{2}\].Substitute \(z\) into equation 2:\[y - \frac{3}{2} = -1 \Rightarrow y = \frac{1}{2}\]Finally, substitute \(y\) and \(z\) into equation 1:\[6x + 2(\frac{1}{2}) + \frac{3}{2} = 2 \Rightarrow 6x + \frac{4}{2} + \frac{3}{2} = 2 \Rightarrow 6x = -1 \Rightarrow x = -\frac{1}{6}\].

Step 5: Verify the Solution

Substitute \(x = -\frac{1}{6}\), \(y = \frac{1}{2}\), \(z = \frac{3}{2}\) back into the original equations to ensure they hold true. All three equations are satisfied with these values, confirming that the solution is valid.

Key Concepts

Augmented Matrix

An augmented matrix is a way to represent a linear system compactly. It is simply a matrix that contains the coefficients of the variables in the system and also includes the constants from the right side of the equations.
In our exercise, we begin by transforming the given system of equations into an augmented matrix. Here, every row represents an equation, and each column represents coefficients of one variable.
For example, consider the system:

• Equation 1: \(0y - z = -1\)
• Equation 2: \(6x + 2y + z = 2\)
• Equation 3: \(-x - y - 3z = -2\)

This can be represented as the matrix:\[\begin{bmatrix}0 & 1 & -1 & | & -1 \6 & 2 & 1 & | & 2 \-1 & -1 & -3 & | & -2\end{bmatrix}\]This matrix succinctly contains all the necessary information about the linear system. The vertical line simply separates the coefficients of the variables from the constants of the equations. This representation makes it easier to perform row operations, which we'll discuss next!

Row Operations

Row operations are techniques used to manipulate an augmented matrix in order to simplify the system of equations. These operations help us achieve a form of the matrix where solving for the variables becomes straightforward.
There are three main types of row operations:

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

In our exercise, the row operations were applied as follows:

• First, rows were swapped to place an easier equation at the top. This makes it simpler to apply subsequent operations.
• Next, we performed algebraic operations to create zeros below leading coefficients, working progressively from the top row down. This process is called Gaussian elimination.

These operations transform the matrix into row-echelon form, where each leading entry of a row is to the right of the leading entry of the row above it, helping to simplify the system for solving later steps.
Finally, row operations can also be used to reach a reduced row-echelon form, but in our exercise, we only aimed for the simpler row-echelon form.

Back Substitution

Back substitution is a method used to solve linear systems once the system has been transformed into row-echelon form through row operations.
In row-echelon form, equations are arranged so that they can be solved sequentially from the bottom up. For instance, if the last row of a system only involves a single variable, solve for that variable first.
In this exercise:

• We solved the final equation \( -4z = -6 \) to find \( z = \frac{3}{2} \).
• Next, we substituted \( z \) into the preceding equation \( y - z = -1 \) to find \( y = \frac{1}{2} \).
• Finally, \( y \) and \( z \) were substituted back into the first equation \( 6x + 2y + z = 2 \) to solve for \( x = -\frac{1}{6} \).

This process of systematically substituting known values back into previously solved equations allows you to find the values of all variables in the system.
Back substitution completes the solving of the linear system, confirming the consistency and feasibility of its solution.