Problem 41 - Algebra for College Students

Question

Describe how to solve the system \(\left(\begin{array}{rl}x-2 y & =-10 \\ 3 x+5 y & =14\end{array}\right)\) using each of the following techniques. (a) substitution method (b) elimination-by-addition method (c) reduced echelon form of the augmented matrix (d) determinants (e) the method of matrix inverses

Step-by-Step Solution

Answer

The solution is \(x = -2\) and \(y = 4\).

Step 1: Introduce the System

The system of linear equations is \( \begin{array}{rl} x-2y & = -10 \ 3x+5y & = 14 \end{array}\). We will solve this using different methods.

Step 2: Substitution Method - Solve for x

From the first equation, solve for \(x\): \(x = 2y - 10\).

Step 3: Substitution Method - Substitute

Substitute the expression for \(x\) into the second equation: \(3(2y - 10) + 5y = 14\).

Step 4: Substitution Method - Simplify and Solve

Simplify and solve for \(y\): \(6y - 30 + 5y = 14\), which simplifies to \(11y = 44\), giving \(y = 4\).

Step 5: Substitution Method - Back Solve for x

Substitute \(y = 4\) back into \(x = 2y - 10\) to find \(x\): \(x = 8 - 10 = -2\).

Step 6: Elimination Method - Align Equations

Start with the original system and multiply the first equation by 3: \(3(x - 2y) = 3(-10)\) giving \(3x - 6y = -30\).

Step 7: Elimination Method - Add Equations

Subtract this new equation from the second original equation: \((3x + 5y) - (3x - 6y) = 14 - (-30)\), giving \(11y = 44\).

Step 8: Elimination Method - Solve for y

From \(11y = 44\), solve for \(y\): \(y = 4\).

Step 9: Elimination Method - Solve for x

Substitute \(y = 4\) into the first equation \(x - 2y = -10\) to solve for \(x\): \(x - 8 = -10\) so \(x = -2\).

Step 10: Augmented Matrix - Setup

Form the augmented matrix from the system: \[ \begin{bmatrix} 1 & -2 & | & -10 \ 3 & 5 & | & 14 \end{bmatrix} \].

Step 11: Augmented Matrix - Row Operations

Use row operations to convert the matrix into reduced row echelon form. First, multiply the first row by 3 and subtract from the second row to eliminate the variables under the first pivot. Solve to get \(y=4\) and backtrack to get \(x=-2\).

Step 12: Determinants - Setup

Recall Cramer's rule requires calculating determinants. Write the coefficient matrix \(A\) and the constant matrix \(B\): \(A = \begin{bmatrix} 1 & -2 \ 3 & 5 \end{bmatrix} \) and \(B = \begin{bmatrix} -10 \ 14 \end{bmatrix}\).

Step 13: Determinants - Calculate Determinants

Calculate \(|A|\) = 1(5) - (-2)(3) = 11. Calculate \(|A_x|\) replacing column of \(x\): = \begin{bmatrix} -10 & -2 \ 14 & 5 \end{bmatrix}, giving \(-10 imes 5) - (-2 imes 14) = -18\) and \(|A_y|\) = \begin{bmatrix} 1 & -10 \ 3 & 14 \end{bmatrix}, giving \(1 \times 14 - 3 \times -10 = 44\).

Step 14: Determinants - Solve Using Cramer's Rule

Now use \(x = \frac{|A_x|}{|A|}\) and \(y = \frac{|A_y|}{|A|}\) to find \(x = \frac{-18}{11} = -2\) and \(y = \frac{44}{11} = 4\).

Step 15: Inverse Matrix - Setup

Write the coefficient matrix \(A\) and vector \(\textbf{b}\): \(A = \begin{bmatrix} 1 & -2 \ 3 & 5 \end{bmatrix}\) and \(\textbf{b} = \begin{bmatrix} -10 \ 14 \end{bmatrix}\).

Step 16: Inverse Matrix - Calculate Inverse of A

Calculate \(A^{-1}\) using the formula: \(A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). Fill values to get: \(A^{-1} = \frac{1}{11} \begin{bmatrix} 5 & 2 \ -3 & 1 \end{bmatrix}\).

Step 17: Inverse Matrix - Solve for x and y

Multiply \(A^{-1}\) by \(\textbf{b}\) to get the solution: \(\begin{bmatrix} x \ y \end{bmatrix} = A^{-1} \cdot \textbf{b}\) resulting in \(\begin{bmatrix} -2 \ 4 \end{bmatrix}\).

Step 18: Summary

Through all methods, the solution to the system is \(x = -2\) and \(y = 4\).

Key Concepts

substitution method

The substitution method is a straightforward technique to solve systems of equations, especially when one equation can be easily solved for one variable.
Here’s how it works:

• First, solve one of the equations for one of the variables in terms of the other. In our example, solve for \(x\) in the equation \(x - 2y = -10\). This gives \(x = 2y - 10\).
• Next, substitute this expression into the other equation. So for the second equation \(3x + 5y = 14\), substitute \(x = 2y - 10\). You get: \(3(2y - 10) + 5y = 14\).
• Simplify and solve for the remaining variable \(y\). In this case, it simplifies to \(11y = 44\) so \(y = 4\).
• Finally, use the value of \(y\) to solve for \(x\) in the expression found earlier: \(x = 2(4) - 10 = -2\).

This method is particularly useful when equations are already set up to make a quick substitution easy.

elimination-by-addition method

The elimination-by-addition method, sometimes called the addition method, is another way to solve systems of equations by eliminating one variable. Here’s the step-by-step approach:

• First, align the equations vertically and manipulate them so that adding or subtracting them eliminates one of the variables. In the example, multiply the first equation by 3 to have the same coefficient of \(x\): \(3x - 6y = -30\).
• Next, add or subtract the equations so that one variable is eliminated. Subtracting the modified first equation from the second yields: \((3x + 5y) - (3x-6y) = 14 - (-30)\), which simplifies to \(11y = 44\).
• Solve for the remaining variable \(y\), which turns out to be \(y = 4\).
• Finally, back substitute \(y\) into one of the original equations to solve for \(x\). Using \(x - 2(4) = -10\), we find \(x = -2\).

This technique is efficient when equations are readily aligned for adding or subtracting and often pairs well with complex systems.

augmented matrix

An augmented matrix is a compact way to represent a system of linear equations. It includes the coefficient matrix alongside the constant matrix. For our problem, the system: \[\begin{array}{rl}x-2y & = -10 \3x+5y & = 14 \\end{array}\]is converted to the format: \[\begin{bmatrix}1 & -2 & | & -10 \3 & 5 & | & 14 \end{bmatrix}\]
To solve using this approach:

• Apply row operations to transform this augmented matrix into reduced row echelon form (RREF). The operations used include swapping rows, multiplying a row by a non-zero scalar, and adding/subtracting rows.
• In the example, eliminate the first variable in the second row by multiplying the first row by 3 and subtracting it from the second row.
• You'll get an equation in terms of \(y\), solve it to find \(y = 4\).
• Then backtrack to solve for \(x\) using \(x -2 (4) = -10\) to find \(x = -2\).

This matrix method is powerful for handling larger systems and automating solutions using computational techniques.

Cramer's rule

Cramer's Rule offers an elegant method to solve systems of linear equations using determinants, perfect for small systems. It's based on the formula involving the determinant of matrices. Here’s how it applies:

• First, construct the coefficient matrix \(A\) and constants matrix \(B\): \[A = \begin{bmatrix}1 & -2 \ 3 & 5 \end{bmatrix}\] and \[B = \begin{bmatrix}-10 \ 14 \end{bmatrix}\]
• Calculate the determinant of \(A\), denoted \(|A|\): \(|A| = 1(5) - (-2)(3) = 11\).
• Then replace each column of \(A\) with \(B\) to find \(|A_x|\) and \(|A_y|\), determinants tailored to solve for \(x\) and \(y\). For \(x\), use matrix: \[\begin{bmatrix}-10 & -2 \ 14 & 5 \end{bmatrix} \Rightarrow |A_x| = -18\]
• Solve for \(x\) and \(y\) using \(x = \frac{|A_x|}{|A|} = -2\) and \(y = \frac{|A_y|}{|A|} = 4\), with \(|A_y| = \begin{bmatrix}1 & -10 \ 3 & 14 \end{bmatrix} = 44\).

Cramer's Rule is efficient but best used when the system size matches the computational capacity.

matrix inverses

The concept of matrix inverses is pivotal in linear algebra. If a square matrix \(A\) has an inverse, denoted as \(A^{-1}\), then the solution to \(Ax = b\) can be found easily: \(x = A^{-1}b\).
Here’s how you use matrix inverses to solve a system:

• Identify the coefficient matrix \(A\): \[A = \begin{bmatrix} 1 & -2 \ 3 & 5 \end{bmatrix}\] and the constant vector \(\textbf{b}\): \[\textbf{b} = \begin{bmatrix} -10 \ 14 \end{bmatrix}\]
• Calculate the inverse of \(A\) if \(\text{det}(A) eq 0\). To find \(A^{-1}\), use the formula based on its determinant: \[A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\] substituting in, get: \[A^{-1} = \frac{1}{11} \begin{bmatrix} 5 & 2 \ -3 & 1 \end{bmatrix}\]
• Multiply \(A^{-1}\) by \(\textbf{b}\) to find the solution vector: \[\begin{bmatrix} x \ y \end{bmatrix} = A^{-1} \cdot \textbf{b} \Rightarrow \begin{bmatrix} -2 \ 4 \end{bmatrix}\]

Using matrix inverses offers a direct solution, especially useful in computational applications and when handling multiple equations.