Problem 54 - 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. $$\begin{aligned} 0.1 x+0.3 y+1.7 z &=0.6 \\ 0.6 x+0.1 y-3.1 z &=6.2 \\ 2.4 y+0.9 z &=3.5 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is approximately \(x = -19.054\), \(y = 1.670\), \(z = 1.179\).

Step 1: Set up the augmented matrix

Start by setting up the augmented matrix of the system. The system of equations is given as: \(\begin{aligned} 0.1x + 0.3y + 1.7z &= 0.6 \ 0.6x + 0.1y - 3.1z &= 6.2 \ 0y + 2.4y + 0.9z &= 3.5\end{aligned}\)The augmented matrix is:\[\left(\begin{array}{ccc|c}0.1 & 0.3 & 1.7 & 0.6 \0.6 & 0.1 & -3.1 & 6.2 \0 & 2.4 & 0.9 & 3.5\end{array}\right)\]

Step 2: Eliminate first column below pivot

Eliminate the first element (0.6) of the second row. Multiply the first row by 6 and subtract it from the second row.New Row 2: \[(0.6)R1 - R2 \Rightarrow 0.6(0.1x + 0.3y + 1.7z = 0.6) - (0.6x + 0.1y - 3.1z = 6.2)\]Updating Row 2:\[0 & -0.08 & -13.46 & -16\]The matrix becomes:\[\left(\begin{array}{ccc|c}0.1 & 0.3 & 1.7 & 0.6 \0 & -0.08 & -13.46 & -16 \0 & 2.4 & 0.9 & 3.5\end{array}\right)\]

Step 3: Eliminate first column in third row

As the first element of the third row is already zero, there's no need for further modifications to the third row in the first column.The matrix remains:\[\left(\begin{array}{ccc|c}0.1 & 0.3 & 1.7 & 0.6 \0 & -0.08 & -13.46 & -16 \0 & 2.4 & 0.9 & 3.5\end{array}\right)\]

Step 4: Scale second row

Scale the second row by dividing it by -0.08 to make the second pivot 1.New Row 2: \[\frac{1}{-0.08}(0)R2 \Rightarrow (0, 1, 168.25, 200)\]The matrix now becomes:\[\left(\begin{array}{ccc|c}0.1 & 0.3 & 1.7 & 0.6 \0 & 1 & 168.25 & 200 \0 & 2.4 & 0.9 & 3.5\end{array}\right)\]

Step 5: Eliminate second column in third row

To eliminate the coefficient of y in the third row, we subtract 2.4 times the new second row from the third row.New Row 3: \[(2.4)R2 - R3 \Rightarrow 0 - (1-(2.4)(0))\]Updating Row 3:\[0 & 0 & -403.8 & -476.5\]The matrix is now:\[\left(\begin{array}{ccc|c}0.1 & 0.3 & 1.7 & 0.6 \0 & 1 & 168.25 & 200 \0 & 0 & -403.8 & -476.5\end{array}\right)\]

Step 6: Simplifying third row

Divide the third row by -403.8 to make the pivot 1.New Row 3: \[\frac{1}{-403.8}(0)R3 \Rightarrow 0 & 0 & 1 & 1.179\]The matrix becomes:\[\left(\begin{array}{ccc|c}0.1 & 0.3 & 1.7 & 0.6 \0 & 1 & 168.25 & 200 \0 & 0 & 1 & 1.179\end{array}\right)\]

Step 7: Back-substitution to find solutions

Substitute the known values back from the last row upwards to find the solutions for y and x.Substituting z = 1.179 into Row 2:\[y + 168.25(1.179) = 200 \y = 200 - 198.32975 = 1.67025\]Substitute y = 1.67025 and z = 1.179 into Row 1:\[0.1x + 0.3(1.67025) + 1.7(1.179) = 0.6 \0.1x = 0.6 - 0.501075 - 2.0043 \0.1x = -1.905375 \x = -19.05375\]The solutions are rounded to the nearest thousandth.

Key Concepts

Augmented Matrix

An augmented matrix is a powerful tool for solving systems of linear equations. It combines the coefficients of the variables and the constants from the equations into a matrix format. This matrix allows us to apply row operations efficiently, simplifying the system step by step.
To create an augmented matrix, line up each equation's coefficients in columns, and the constants on the right in a vertical bar format. For example, the equation system given as:

• \(0.1x + 0.3y + 1.7z = 0.6\)
• \(0.6x + 0.1y - 3.1z = 6.2\)
• \(2.4y + 0.9z = 3.5\)

Transforms into the augmented matrix:\[\left(\begin{array}{ccc|c}0.1 & 0.3 & 1.7 & 0.6 \0.6 & 0.1 & -3.1 & 6.2 \0 & 2.4 & 0.9 & 3.5\end{array}\right)\] This form enables the use of row operations to simplify the matrix and solve for the variables.

Systems of Linear Equations

A system of linear equations consists of two or more equations that share variables. Solving these systems means finding the values of the variables that satisfy all the equations simultaneously. In our exercise, we have three linear equations with three variables: \(x\), \(y\), and \(z\).
There are several methods to solve systems of linear equations, such as substitution, elimination, and using matrices. Each method has its strengths, and augmented matrices are particularly effective for larger systems because they simplify complex operations.
When working with an augmented matrix, you aim to manipulate it into what is called row echelon form. This form makes solving the system attainable through straightforward back-substitution. For instance, after performing row operations, our augmented matrix steps closer to this ideal structure, simplifying the solution process.

Back-substitution

Back-substitution is the final step in solving a system of equations using an augmented matrix. Once the matrix is in row echelon form, we can easily substitute values starting from the last row to find the solution. This process involves determining the value of one variable at a time, going backwards from the last to the first row.
In our solution, the last row simplifies to \(z = 1.179\). With \(z\) known, we substitute it into the second row equation to solve for \(y\). The equation \(y + 168.25z = 200\) simplifies to \(y = 1.67025\) as \(z\) gets substituted.
The final substitution happens in the first row, where both \(y\) and \(z\) are known. Plugging these values, the equation becomes \(0.1x = -1.905375\), leading to \(x = -19.054\) after rounding to the nearest thousandth. Thus, each step builds on the previous, ensuring each solution is accurate and complete.