Problem 49 - Precalculus: Mathematics for Calculus

Question

Use Cramer's Rule to solve the system. $$\left\\{\begin{aligned} 2 x_{1}+3 x_{2}-5 x_{3} &=1 \\ x_{1}+x_{2}-x_{3} &=2 \\ 2 x_{2}+x_{3} &=8 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \( x_1 = 1 \), \( x_2 \approx 8.71 \), \( x_3 = 2 \).

Step 1: Write the System in Matrix Form

First, express the given system of equations in matrix form as \( AX = B \). For the system: \[\begin{align*}2x_1 + 3x_2 - 5x_3 &= 1 \x_1 + x_2 - x_3 &= 2 \0x_1 + 2x_2 + x_3 &= 8\end{align*}\]The matrix \( A \) is:\[\begin{bmatrix}2 & 3 & -5 \1 & 1 & -1 \0 & 2 & 1\end{bmatrix}\]The matrix \( X \) (containing the variables) is:\[\begin{bmatrix}x_1 \x_2 \x_3\end{bmatrix}\]And \( B \) (the constants) is:\[\begin{bmatrix}1 \2 \8\end{bmatrix}\]

Step 2: Calculate the Determinant of A

Find \( \text{det}(A) \) for the matrix \( A \). The determinant is calculated as follows for a 3x3 matrix:\[\text{det}(A) = 2\begin{vmatrix} 1 & -1 \ 2 & 1 \end{vmatrix} - 3\begin{vmatrix} 1 & -1 \ 0 & 1 \end{vmatrix} - 5\begin{vmatrix} 1 & 1 \ 0 & 2 \end{vmatrix}\]Calculate each 2x2 determinant:\[\begin{vmatrix} 1 & -1 \ 2 & 1 \end{vmatrix} = (1 \times 1) - (-1 \times 2) = 1 + 2 = 3\]\[\begin{vmatrix} 1 & -1 \ 0 & 1 \end{vmatrix} = (1 \times 1) - (-1 \times 0) = 1 - 0 = 1\]\[\begin{vmatrix} 1 & 1 \ 0 & 2 \end{vmatrix} = (1 \times 2) - (1 \times 0) = 2\]Substitute and compute:\[\text{det}(A) = 2(3) - 3(1) - 5(2) = 6 - 3 - 10 = -7\]

Step 3: Calculate Determinants for Cramer's Rule

Next, calculate \( \text{det}(A_1) \), \( \text{det}(A_2) \), \( \text{det}(A_3) \), by replacing the appropriate column in \( A \) with \( B \). For \( \text{det}(A_1) \), replace the first column of \( A \) with \( B \):\[A_1 = \begin{bmatrix}1 & 3 & -5 \2 & 1 & -1 \8 & 2 & 1\end{bmatrix}\]\[\text{det}(A_1) = 1\begin{vmatrix} 1 & -1 \ 2 & 1 \end{vmatrix} - 3\begin{vmatrix} 2 & -1 \ 8 & 1 \end{vmatrix} - 5\begin{vmatrix} 2 & 1 \ 8 & 2 \end{vmatrix}\]Calculate each:\[\begin{vmatrix} 1 & -1 \ 2 & 1 \end{vmatrix} = 3\]\[\begin{vmatrix} 2 & -1 \ 8 & 1 \end{vmatrix} = (2 \times 1) - (-1 \times 8) =2 + 8 = 10\]\[\begin{vmatrix} 2 & 1 \ 8 & 2 \end{vmatrix} = (2 \times 2) - (1 \times 8) = 4 - 8 = -4\]Substitute to find \( \text{det}(A_1) \):\[\text{det}(A_1) = 1(3) - 3(10) - 5(-4) = 3 - 30 + 20 = -7\]

Step 4: Continue Determinant Calculations

Compute \( \text{det}(A_2) \), where we replace the second column:\[A_2 = \begin{bmatrix}2 & 1 & -5 \1 & 2 & -1 \0 & 8 & 1\end{bmatrix}\]\[\text{det}(A_2) = 2\begin{vmatrix} 2 & -1 \ 8 & 1 \end{vmatrix} - 1\begin{vmatrix} 1 & -1 \ 0 & 1 \end{vmatrix} - 5\begin{vmatrix} 1 & 2 \ 0 & 8 \end{vmatrix}\]Calculate:\[\begin{vmatrix} 2 & -1 \ 8 & 1 \end{vmatrix} = -10\]\[\begin{vmatrix} 1 & -1 \ 0 & 1 \end{vmatrix} = 1\]\[\begin{vmatrix} 1 & 2 \ 0 & 8 \end{vmatrix} = 8\]Substitute to find \( \text{det}(A_2) \):\[\text{det}(A_2) = 2(-10) - 1(1) - 5(8) = -20 - 1 - 40 = -61\]

Step 5: Final Determinant Calculation

Calculate \( \text{det}(A_3) \) by replacing the third column:\[A_3 = \begin{bmatrix}2 & 3 & 1 \1 & 1 & 2 \0 & 2 & 8\end{bmatrix}\]\[\text{det}(A_3) = 2\begin{vmatrix} 1 & 2 \ 2 & 8 \end{vmatrix} - 3\begin{vmatrix} 1 & 2 \ 0 & 8 \end{vmatrix} + 1\begin{vmatrix} 1 & 1 \ 0 & 2 \end{vmatrix}\]Calculate:\[\begin{vmatrix} 1 & 2 \ 2 & 8 \end{vmatrix} = 4\]\[\begin{vmatrix} 1 & 2 \ 0 & 8 \end{vmatrix} = 8\]\[\begin{vmatrix} 1 & 1 \ 0 & 2 \end{vmatrix} = 2\]Substitute to find \( \text{det}(A_3) \):\[\text{det}(A_3) = 2(4) - 3(8) + 1(2) = 8 - 24 + 2 = -14\]

Step 6: Apply Cramer's Rule for Solutions

Use Cramer's Rule to find \( x_1, x_2, \text{ and } x_3 \). Cramer's Rule states:\[x_1 = \frac{\text{det}(A_1)}{\text{det}(A)}, \quad x_2 = \frac{\text{det}(A_2)}{\text{det}(A)}, \quad x_3 = \frac{\text{det}(A_3)}{\text{det}(A)}\]Substituting the determinants:\[x_1 = \frac{-7}{-7} = 1, \quad x_2 = \frac{-61}{-7} \approx 8.71, \quad x_3 = \frac{-14}{-7} = 2\]

Step 7: Conclusion

The solution to the system using Cramer's Rule is \( x_1 = 1 \), \( x_2 \approx 8.71 \), and \( x_3 = 2 \). This means these are the values that satisfy all the given equations.

Key Concepts

Determinants

When diving into matrix algebra, determinants stand out as crucial elements, especially when using Cramer's Rule to solve systems of equations. A determinant is a special number that can be calculated from a square matrix, and it provides scalar information about the matrix. In a 3x3 matrix, like the one in this exercise, calculating the determinant involves creating and adding up specific products of its elements. Here's a simple way to think about determinants:

• The determinant reflects certain properties of a matrix, such as invertibility. If the determinant is not zero, the matrix has an inverse.
• The magnitude of the determinant can give insight into the geometric properties of the matrix transformation, such as scaling a space.
• Determinants are pivotal for solving linear equations because they determine the solvability and behavior of a system of equations.

In the given exercise, determinants help us find solutions by applying Cramer's Rule. Every step, from finding \(\text{det}(A)\) for the original coefficient matrix to calculating \(\text{det}(A_1)\), \(\text{det}(A_2)\), and \(\text{det}(A_3)\) (by replacing columns with the constant matrix \(B\)), revolves around these determinant calculations to achieve the solution.

Matrix Algebra

Matrix algebra is a branch of mathematics that deals with matrices and the operations that can be performed on them. It enables us to solve complex systems of linear equations efficiently. Here are some key concepts:

• In matrix algebra, we can multiply matrices, add them, and calculate determinants, among other operations. Each operation has specific rules.


• Matrix multiplication is not like regular multiplication; it involves dot products of rows and columns, leading to a new matrix.


• Matrix algebra is essential for representing systems of equations. By using matrices, we can rewrite and simplify large systems, making them easier to manipulate and solve.

In our particular exercise, the system of linear equations was first expressed in matrix form as \(AX = B\). Here, \(A\) represents the coefficients, \(X\) is the vector of variables, and \(B\) denotes the constants. This setup allows us to utilize various matrix operations to systematically approach and solve the system using techniques like Cramer's Rule.

System of Equations

A system of equations is comprised of multiple linear equations that are solved simultaneously. Here’s why they are important and how Cramer's Rule applies:

• Systems of equations often model real-world problems, such as in physics or economics, where several variables are interdependent.
• To solve such systems, one aims to find values for all variables that satisfy every equation in the system. This simultaneous solving is what makes systems of equations powerful.


• Cramer's Rule is a method used when the system of equations can be written in the matrix form \(AX = B\), assuming that \(A\) is a square matrix with a non-zero determinant. It uses determinants to find the unique solution for the variables if it exists.

By using Cramer's Rule, as shown in the exercise, we transform the initially complicated process of solving simultaneous equations into a series of determinant calculations, which are straightforward and computationally manageable steps. Each variable \(x_1, x_2, x_3\) has a solution that stems from this neat application of matrix algebra.