Problem 26 - Elementary and Intermediate Algebra

Question

Solve each system. If a system’s equations are dependent or if there is no solution, state this. $$\begin{aligned} 5 x+3 y+\frac{1}{2} z &=\frac{7}{2} ,\\\ 0.5 x-0.9 y-0.2 z &=0.3 ,\\\ 3 x-2.4 y+0.4 z &=-1 \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system is \(x, y, z\) = (x, \frac{11}{224}, -\frac{195}{920}\).

Step 1 - Write the system in matrix form

Write the given system of equations in augmented matrix form:\[ \begin{pmatrix} 5 & 3 & \frac{1}{2} & \bigg| & \frac{7}{2} \ 0.5 & -0.9 & -0.2 & \bigg| & 0.3 \ 3 & -2.4 & 0.4 & \bigg| & -1 \end{pmatrix} \]

Step 2 - Convert to a consistent format

Convert all fractions and decimals to a consistent format before performing row operations. For simplicity, multiply the entire matrix by 2 to clear the \(\frac{1}{2}\):\[ \begin{pmatrix} 10 & 6 & 1 & \bigg| & 7 \ 1 & -1.8 & -0.4 & \bigg| & 0.6 \ 6 & -4.8 & 0.8 & \bigg| & -2 \end{pmatrix} \]

Step 3 - Perform row operations to create zeros below the leading 1 in the first column

Let \(R_2 = R_2 - 0.1R_1\) and \(R_3 = R_3 - 0.6R_1\):\[ \begin{pmatrix} 10 & 6 & 1 & \bigg| & 7 \ 0 & -2.4 & -0.5 & \bigg| & -0.1 \ 0 & -8.4 & 0.2 & \bigg| & -6.2 \end{pmatrix} \]

Step 4 - Simplify Row 2

Divide Row 2 by -2.4 to simplify:\[ \begin{pmatrix} 10 & 6 & 1 & \bigg| & 7 \ 0 & 1 & \frac{5}{12} & \bigg| & \frac{1}{24} \ 0 & -8.4 & 0.2 & \bigg| & -6.2 \end{pmatrix} \]

Step 5 - Eliminate the y term from Row 3

Let \(R_3 = R_3 + 8.4R_2\):\[ \begin{pmatrix} 10 & 6 & 1 & \bigg| & 7 \ 0 & 1 & \frac{5}{12} & \bigg| & \frac{1}{24} \ 0 & 0 & 4.6 & \bigg| & -4.875 \end{pmatrix} \]

Step 6 - Simplify Row 3

Divide Row 3 by 4.6:\[ \begin{pmatrix} 10 & 6 & 1 & \bigg| & 7 \ 0 & 1 & \frac{5}{12} & \bigg| & \frac{1}{24} \ 0 & 0 & 1 & \bigg| & -\frac{195}{920} \end{pmatrix} \]

Step 7 - Perform back substitution

Perform back substitution starting from the last row:\(z = -\frac{195}{920}\)\[ \begin{aligned} y + \frac{5}{12}z &= \frac{1}{24} \ y + \frac{5}{12}(-\frac{195}{920}) &= \frac{1}{24} \ y - \frac{975}{11040} &= \frac{1}{24} \ y &= \frac{495}{11040} \ y &= \frac{11}{224} \end{aligned} \] Finally, substitute \y\ and \z\ into the first equation to find \x\.

Step 8 - Find the value of x

Substitute \y = \frac{11}{224}\ and \z = -\frac{195}{920}\ into the first equation:\[ \begin{aligned} 10x + 6(\frac{11}{224}) + (-\frac{195}{920}) &= 7 \end{aligned} \] Solve for \x\.

Key Concepts

Matrix Representation

To solve complex systems of equations, one effective method is to use matrix representation. This involves rewriting the system as an augmented matrix, where each row represents an equation and each column represents the coefficients of the variables. For instance, consider the following system of equations:
\[ \begin{aligned} 5x + 3y + \frac{1}{2} z &= \frac{7}{2} \ 0.5x - 0.9y - 0.2z &= 0.3 \ 3x - 2.4y + 0.4z &= -1 \end{aligned} \] In augmented matrix form, this becomes:
\[ \begin{pmatrix} 5 & 3 & \frac{1}{2} & \bigg| & \frac{7}{2} \ 0.5 & -0.9 & -0.2 & \bigg| & 0.3 \ 3 & -2.4 & 0.4 & \bigg| & -1 \end{pmatrix} \] This compact form is much easier to manipulate using matrix row operations.

Row Operations

Row operations are manipulations you perform on the rows of the matrix to simplify it and solve the system. These operations fall into three categories: swapping two rows, multiplying a row by a non-zero constant, and adding or subtracting a multiple of one row to/from another row. In our example:
\[ \begin{pmatrix} 5 & 3 & \frac{1}{2} & \bigg| & \frac{7}{2} \ 0.5 & -0.9 & -0.2 & \bigg| & 0.3 \ 3 & -2.4 & 0.4 & \bigg| & -1 \end{pmatrix} \] First, to clear fractions and decimals, you might multiply the whole matrix by 2:
\[ \begin{pmatrix} 10 & 6 & 1 & \bigg| & 7 \ 1 & -1.8 & -0.4 & \bigg| & 0.6 \ 6 & -4.8 & 0.8 & \bigg| & -2 \end{pmatrix} \] Then, apply row operations to create zeros below leading ones, simplifying systematically:
\[ \begin{pmatrix} 10 & 6 & 1 & \bigg| & 7 \ 0 & -2.4 & -0.5 & \bigg| & -0.1 \ 0 & -8.4 & 0.2 & \bigg| & -6.2 \end{pmatrix} \]

Back Substitution

After transforming the matrix into an upper triangular form, back substitution is used to solve for the variables one at a time, starting from the last row. For example, a simplified matrix could look like this:
\[ \begin{pmatrix} 10 & 6 & 1 & \bigg| & 7 \ 0 & 1 & \frac{5}{12} & \bigg| & \frac{1}{24} \ 0 & 0 & 1 & \bigg| & -\frac{195}{920} \end{pmatrix} \] Here, you can immediately solve for \( z \) in the last row: \( z = -\frac{195}{920} \). Substitute \( z \) into the second row:
\[ y + \frac{5}{12}(-\frac{195}{920}) = \frac{1}{24} \] Solving this, you get \( y = \frac{11}{224} \). Finally, substitute both \( y \) and \( z \) into the first row to get \( x \):
\[ 10x + 6(\frac{11}{224}) + (-\frac{195}{920}) = 7 \] Simplifying yields the value of \( x \).

Consistency in Formats

Whenever solving systems of equations using matrices, maintaining consistency in formats is crucial. Convert all fractions and decimals to a single consistent format to prevent errors and simplify calculations. In our example, eliminating the fractions and decimals involved multiplying the entire matrix by 2:
\[ \begin{pmatrix} 10 & 6 & 1 & \bigg| & 7 \ 1 & -1.8 & -0.4 & \bigg| & 0.6 \ 6 & -4.8 & 0.8 & \bigg| & -2 \end{pmatrix} \] Working with whole numbers and keeping consistent format ensures operations are error-free and the process is smoother. This consistency is important in both manual calculations and when using computational tools.