Problem 46 - A Graphical Approach to College Algebra

Question

Use row operations on an augmented matrix to solve each system of equations. Round to nearest thousandth when appropriate. $$\begin{aligned} x+y &=1 \\ 2 x-z &=0 \\ y+2 z &=-2 \end{aligned}$$

Step-by-Step Solution

Answer

Solution: \(x = 1\), \(y = 2\), \(z = -2\).

Step 1: Set up the augmented matrix

First, write the system of equations as an augmented matrix. The system is: \(x + y = 1\), \(2x - z = 0\), and \(y + 2z = -2\). The corresponding augmented matrix is:\[\begin{bmatrix}1 & 1 & 0 & | & 1 \2 & 0 & -1 & | & 0 \0 & 1 & 2 & | & -2 \end{bmatrix}\]

Step 2: Use row operations to get a leading 1 in first row, first column

The first row already has a leading 1 in the first column, so no changes are needed here. The matrix remains:\[\begin{bmatrix}1 & 1 & 0 & | & 1 \2 & 0 & -1 & | & 0 \0 & 1 & 2 & | & -2\end{bmatrix}\]

Step 3: Eliminate the first column in the second row

Subtract 2 times the first row from the second row to eliminate the first element in the second row:\(R_2 = R_2 - 2R_1\)\[\begin{bmatrix}1 & 1 & 0 & | & 1 \0 & -2 & -1 & | & -2 \0 & 1 & 2 & | & -2\end{bmatrix}\]

Step 4: Make a leading 1 in the second row, second column

Divide the second row by -2 to have a leading 1 in the second row:\(R_2 = \frac{-1}{2}R_2\)\[\begin{bmatrix}1 & 1 & 0 & | & 1 \0 & 1 & \frac{1}{2} & | & 1 \0 & 1 & 2 & | & -2\end{bmatrix}\]

Step 5: Eliminate the second column in the first and third rows

Subtract the second row from the first row to eliminate the first row's second element: \(R_1 = R_1 - R_2\), and subtract the second row from the third row: \(R_3 = R_3 - R_2\).For the first row:\[1 - 0 = 1, \quad 1 - 1 = 0, \quad 0 - \frac{1}{2} = -\frac{1}{2}, \quad 1 - 1 = 0\]For the third row:\[0 - 0 = 0, \quad 1 - 1 = 0, \quad 2 - \frac{1}{2} = \frac{3}{2}, \quad -2 - 1 = -3\]Now the matrix is:\[\begin{bmatrix}1 & 0 & -\frac{1}{2} & | & 0 \0 & 1 & \frac{1}{2} & | & 1 \0 & 0 & \frac{3}{2} & | & -3\end{bmatrix}\]

Step 6: Solve for the third row

Divide the third row by \(\frac{3}{2}\) to solve for \(z\):\[0 = 0, \quad 0 = 0, \quad \frac{3}{2}z = -3\]\(z = \frac{-3}{\frac{3}{2}} = -2\).Now update the matrix with this value:\[\begin{bmatrix}1 & 0 & -\frac{1}{2} & | & 0 \0 & 1 & \frac{1}{2} & | & 1 \0 & 0 & 1 & | & -2\end{bmatrix}\]

Step 7: Back-substitute to find \(y\) and \(x\)

Substitute \(z = -2\) into the second row to solve for \(y\):\[0 \cdot 0 + y + \frac{1}{2}(-2) = 1 \y - 1 = 1 \y = 2\]Substitute \(z = -2\) into the first row to solve for \(x\):\[x = 0 + \frac{1}{2}(2) + 0 \0 = x - 1 \x = 1\]

Step 8: Conclude the solution

With the obtained values, the solution to the system of equations is \(x = 1\), \(y = 2\), and \(z = -2\). We used Gaussian elimination to find these values by transforming the augmented matrix into reduced row-echelon form.

Key Concepts

Row Operations

In solving systems of equations using matrices, row operations are vital. These operations help transform the matrix into a form that is easier to solve. Here are the main types of row operations you can use:

• Swapping two rows: This operation involves exchanging positions of any two rows in the matrix.
• Multiplying a row by a non-zero constant: This allows us to adjust the values of a row to make calculations easier, like achieving leading 1s.
• Adding or subtracting rows: You can add or subtract one row from another to eliminate coefficients or constants. This helps create zeros in strategic positions.

Applying these operations step-by-step can simplify complex systems and is crucial in methods like Gaussian elimination.

Gaussian Elimination

Gaussian elimination is a systematic method of using row operations to solve systems of linear equations. By converting the system's augmented matrix into a more usable form, it makes finding solutions straightforward. The process involves:

• Creating a leading 1 in the row of the matrix (pivoting step).
• Using that leading 1 to eliminate all other values in the same column, both above and below (elimination step).

Repeat these steps systematically for each column in the matrix. The primary goal is to simplify the system step-by-step until you arrive at a triangular form, which will later convert to a reduced row-echelon form.

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In the context of linear algebra, these equations can be solved using matrices and methods like Gaussian elimination. The purpose of solving such a system is to find the value of the variables that satisfies all given equations simultaneously.
For the given example, you have three equations with three variables: \(x, y,\) and \(z\). The aim is to determine the values of \(x, y,\) and \(z\) where all three equations are true at the same time. This solution process typically involves organizing the equations into an augmented matrix and systematically applying row operations.

Reduced Row-Echelon Form

Reduced row-echelon form (RREF) is the goal of using Gaussian elimination. It is a matrix form where all leading coefficients (the first non-zero number from the left in each row) are 1, and each leading 1 is the only non-zero entry in its column. Additionally, each leading 1 appears to the right of the leading 1 in the row above it. The advantages of RREF include:

• Matrices in RREF make it very easy to identify solutions to the system of equations.
• The solutions can be read directly from the matrix once it is in RREF.

Achieving RREF involves eliminating unnecessary numbers through row operations until all conditions of RREF are met. In the example above, by transforming the augmented matrix to RREF, you can directly see that \(x = 1\), \(y = 2\), and \(z = -2\). This systematic approach makes solving linear systems efficient and understandable.