Problem 33 - Precalculus Mathematics for Calculus

Question

Use Cramer's Rule to solve the system. $$\left\\{\begin{array}{rr} 2 x-y= & -9 \\ x+2 y= & 8 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = -2\) and \(y = 5\).

Step 1: Write the system as a matrix equation

The given system of equations can be expressed in the matrix form: \[\begin{bmatrix} 2 & -1 \ 1 & 2 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -9 \ 8 \end{bmatrix}\] Here, the coefficient matrix \(A\) is \(\begin{bmatrix} 2 & -1 \ 1 & 2 \end{bmatrix}\), and the constants matrix is \(\begin{bmatrix} -9 \ 8 \end{bmatrix}\).

Step 2: Calculate the determinant of the coefficient matrix (D)

The determinant \(D\) of matrix \(A\) is calculated as follows:\[D = \begin{vmatrix} 2 & -1 \ 1 & 2 \end{vmatrix} = (2)(2) - (-1)(1) = 4 + 1 = 5\]So, \(D = 5\).

Step 3: Replace the first column to find the determinant for x (Dx)

Replace the first column of the coefficient matrix \(A\) with the constants from the equations matrix:\[A_x = \begin{bmatrix} -9 & -1 \ 8 & 2 \end{bmatrix}\]Calculate the determinant \(D_x\):\[D_x = \begin{vmatrix} -9 & -1 \ 8 & 2 \end{vmatrix} = (-9)(2) - (-1)(8) = -18 + 8 = -10\]

Step 4: Replace the second column to find the determinant for y (Dy)

Replace the second column of the coefficient matrix \(A\) with the constants from the equations matrix:\[A_y = \begin{bmatrix} 2 & -9 \ 1 & 8 \end{bmatrix}\]Calculate the determinant \(D_y\):\[D_y = \begin{vmatrix} 2 & -9 \ 1 & 8 \end{vmatrix} = (2)(8) - (1)(-9) = 16 + 9 = 25\]

Step 5: Use Cramer's Rule to solve for x and y

Using Cramer's Rule, solve for \(x\) and \(y\):\[x = \frac{D_x}{D} = \frac{-10}{5} = -2\]\[y = \frac{D_y}{D} = \frac{25}{5} = 5\]Therefore, the solution to the system is \(x = -2\) and \(y = 5\).

Key Concepts

Determinant Calculation

When solving linear systems using Cramer's Rule, calculating the determinant is crucial. The determinant is a special number that can be calculated from a square matrix. For a 2x2 matrix, it's quite simple. You multiply the top left by the bottom right and then subtract the product of the top right and bottom left.
Let's break down this process using our example coefficient matrix:

• Given matrix is \[\begin{bmatrix} 2 & -1 \ 1 & 2 \end{bmatrix}\]
• Calculate the determinant \(D\): \[ (2 \times 2) - (-1 \times 1) = 4 + 1 = 5 \]

This calculated determinant, \(D = 5\), tells us important information about the matrix. Specifically, since \(D eq 0\), this matrix is invertible and a unique solution exists.
Determinants have this ability to signal whether a solution is possible, which is why they are a key component in Cramer's Rule.

Matrix Equations

Understanding how to express a system of equations as a matrix equation is a foundational skill in linear algebra. It allows the use of matrix operations to solve the system.
In our exercise, the system of equations is:

• \[2x - y = -9 \]
• \[ x + 2y = 8 \]

This corresponds to a matrix equation: \[\begin{bmatrix} 2 & -1 \ 1 & 2 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -9 \ 8 \end{bmatrix}\]Here:

• The first matrix is the **coefficient matrix**, consisting of the coefficients of the variables \(x\) and \(y\).
• The second matrix is the **variable matrix**, containing our unknowns, \(x\) and \(y\).
• The last matrix is the **constants matrix**.

Writing equations in this form simplifies the use of various mathematical procedures, such as applying Cramer's Rule or performing operations like matrix inversion.

Solving Linear Systems

Solving linear systems, especially with multiple variables, often involves sophisticated approaches beyond simple substitution or elimination. Cramer's Rule is a method specifically designed for systems where the same number of equations and unknowns exist.
To apply Cramer's Rule, you need to:

• Calculate the determinant of the original coefficient matrix.
• Create new matrices by replacing one column at a time with the constants matrix, while keeping the other columns unchanged.
• Find the determinant of each modified matrix.
• Divide the determinant of each modified matrix by the determinant of the original matrix.

In our exercise:

• For \(x\), replace the first column of the coefficient matrix with the constants to find \(D_x\). Calculate \(D_x = -10\).
• For \(y\), replace the second column to find \(D_y\). Calculate \(D_y = 25\).
• Solve for \(x\) as \(\frac{D_x}{D} = \frac{-10}{5} = -2\).
• Solve for \(y\) as \(\frac{D_y}{D} = \frac{25}{5} = 5\).

This method efficiently finds solutions, assuming the determinant of the coefficient matrix is non-zero, confirming the matrix's invertibility and the existence of a unique solution.