Problem 47 - 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. $$\begin{aligned} 2 x-y+3 z &=0 \\ x+2 y-z &=5 \\ 2 y+z &=1 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = -0.667\), \(y = 3.667\), \(z = 1.667\).

Step 1: Write the System as an Augmented Matrix

The given system of equations can be represented as an augmented matrix. Each row of the matrix corresponds to an equation in the system:\[\begin{bmatrix}2 & -1 & 3 & | & 0 \1 & 2 & -1 & | & 5 \0 & 2 & 1 & | & 1 \\end{bmatrix}\]

Step 2: Swap Rows to Simplify the Matrix

To simplify operations, we can begin by swapping the first row with the second row for computational convenience:\[\begin{bmatrix}1 & 2 & -1 & | & 5 \2 & -1 & 3 & | & 0 \0 & 2 & 1 & | & 1 \\end{bmatrix}\]

Step 3: Eliminate the First Variable from the Second Row

We want to eliminate the variable \(x\) from the second row. We can do this by subtracting 2 times the first row from the second row:2(R1) from R2:\(R2 = R2 - 2R1\)\[\begin{bmatrix}1 & 2 & -1 & | & 5 \0 & -5 & 5 & | & -10 \0 & 2 & 1 & | & 1 \\end{bmatrix}\]

Step 4: Eliminate the First Variable from the Third Row

No changes are necessary for the third row in terms of the first variable since it already has zero coefficient for \(x\). So the matrix remains:\[\begin{bmatrix}1 & 2 & -1 & | & 5 \0 & -5 & 5 & | & -10 \0 & 2 & 1 & | & 1 \\end{bmatrix}\]

Step 5: Eliminate the Second Variable from the Third Row

To eliminate \(y\) in the third row, we add \(\frac{2}{-5}\) of R2 to R3:\(R3 = R3 + \frac{2}{-5}R2\)\[\begin{bmatrix}1 & 2 & -1 & | & 5 \0 & -5 & 5 & | & -10 \0 & 0 & 3 & | & 5 \\end{bmatrix}\]

Step 6: Solve for the Third Variable

With the matrix in triangular form, solve for \(z\) in the third equation:\(3z = 5\) Thus, \(z = \frac{5}{3} = 1.667\) (rounded to the nearest thousandth).

Step 7: Back Substitution to Solve for the Second Variable

Use the value of \(z\) to find \(y\):From the second row equation:\(-5y + 5z = -10\)Substitute \(z = 1.667\):\(-5y + 5(1.667) = -10\)\(-5y + 8.335 = -10\)\(-5y = -18.335\)\(y = 3.667\) (rounded to the nearest thousandth).

Step 8: Solve for the First Variable

Finally, substitute \(y = 3.667\) and \(z = 1.667\) back into the first equation to solve for \(x\):\(x + 2y - z = 5\)\(x + 2(3.667) - 1.667 = 5\)\(x + 7.334 - 1.667 = 5\)\(x + 5.667 = 5\)\(x = -0.667\) (rounded to the nearest thousandth).

Key Concepts

Row Operations

Row operations are powerful techniques used to manipulate matrices in order to simplify systems of equations. They are essential in transforming a system into a form that is easier to solve. There are three main types of row operations:

• Swapping two rows.
• Multiplying a row by a nonzero scalar.
• Adding or subtracting the multiple of one row from another row.

These operations preserve the solution of the system while making it more manageable. In our case, row operations are used to turn our augmented matrix into a simpler triangular form. By strategically applying these operations, we can ensure that each step reduces the complexity of the system.

Systems of Equations

A system of equations is a set of two or more equations with shared variables. These systems can appear in many real-world problems, and solving them often requires a solid understanding of algebraic principles. The goal is to find values for the variables that make all the equations true simultaneously. In matrix form, systems of equations can be represented as augmented matrices. Each row corresponds to an equation, and each column to a variable. The last column represents the constants. This compact representation helps visualize and manipulate the system to find solutions. The system given in our exercise was written as:\[\begin{bmatrix}2 & -1 & 3 & | & 0 \1 & 2 & -1 & | & 5 \0 & 2 & 1 & | & 1 \end{bmatrix}\]

Back Substitution

Back substitution is the method used to solve equations from the bottom up once the system has been reduced to triangular form. After solving for the last variable, you work your way upwards, substituting each solved variable into the previous equations.For example, once you have solved for the last variable, say \(z\), from the last row of the matrix, you will plug that value into the second-to-last equation to solve for the next variable, \(y\). With \(y\) and \(z\) known, substitute these values into the first row equation to find \(x\). Each step simplifies the process as you carry known values backwards through the system.This method allows for an orderly, step-by-step solution of complex systems, making it much more straightforward.

Triangular Form

Triangular form refers to the transformation of our augmented matrix such that all entries below the main diagonal are zeros. This configuration is crucial because it allows for straightforward back substitution.A matrix is said to be in upper triangular form when it has non-zero elements along the diagonal or above it, and all elements below the diagonal are zero. For example, in our final matrix:\[\begin{bmatrix}1 & 2 & -1 & | & 5 \0 & -5 & 5 & | & -10 \0 & 0 & 3 & | & 5 \end{bmatrix}\]Upper triangular form clears the path for solving the system using back substitution, as shown in the original problem. This form simplifies the system, reducing it to the solvable individual equations of each variable, one at a time.