Problem 46 - Algebra and Trigonometry

Question

Use Cramer’s Rule to solve the system. $$ \left\\{\begin{aligned} 2 x-5 y =4 \\ x+y-z =8 \\ 3 x +5 z=0 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{2}{5}, y = \frac{24}{25}, z = -\frac{84}{25} \).

Step 1: Identify the Coefficient Matrix

The coefficient matrix for the given system of equations is \[ A = \begin{bmatrix} 2 & -5 & 0 \ 1 & 1 & -1 \ 3 & 0 & 5 \end{bmatrix} \]

Step 2: Compute the Determinant of A

Calculate the determinant of matrix \( A \) using the formula for a 3x3 matrix:\[\det(A) = 2 \cdot \begin{vmatrix} 1 & -1 \ 0 & 5 \end{vmatrix} - (-5) \cdot \begin{vmatrix} 1 & -1 \ 3 & 5 \end{vmatrix} + 0 \cdot \begin{vmatrix} 1 & 1 \ 3 & 0 \end{vmatrix} \]\[= 2(1 \cdot 5 - (-1) \cdot 0) + 5(1 \cdot 5 - (-1) \cdot 3)\]\[= 2(5) + 5(5 + 3) = 10 + 5(8) = 10 + 40 = 50\]

Step 3: Formulate Matrices for x, y, and z

Replace the corresponding columns in matrix \( A \) with the constant matrix \( B = \begin{bmatrix} 4 \ 8 \ 0 \end{bmatrix} \) to form matrices \( A_x, A_y, A_z \):\( A_x = \begin{bmatrix} 4 & -5 & 0 \ 8 & 1 & -1 \ 0 & 0 & 5 \end{bmatrix} \),\( A_y = \begin{bmatrix} 2 & 4 & 0 \ 1 & 8 & -1 \ 3 & 0 & 5 \end{bmatrix} \),\( A_z = \begin{bmatrix} 2 & -5 & 4 \ 1 & 1 & 8 \ 3 & 0 & 0 \end{bmatrix} \)

Step 4: Calculate Determinants for A_x, A_y, A_z

Compute the determinants:\[\det(A_x) = 4 \cdot \begin{vmatrix} 1 & -1 \ 0 & 5 \end{vmatrix} - (-5) \cdot \begin{vmatrix} 8 & -1 \ 0 & 5 \end{vmatrix}\]\[= 4(5) + 5(40) = 20 + 0 = 20\]\[\det(A_y) = 2 \cdot \begin{vmatrix} 8 & -1 \ 0 & 5 \end{vmatrix} - 4 \cdot \begin{vmatrix} 1 & -1 \ 3 & 5 \end{vmatrix}\]\[= 2(40) - 4(5 + 3) = 80 - 4(8) = 80 - 32 = 48\]\[\det(A_z) = 2 \cdot \begin{vmatrix} 1 & 8 \ 3 & 0 \end{vmatrix} - (-5) \cdot \begin{vmatrix} 1 & 8 \ 3 & 0 \end{vmatrix}\]\[= 2(-24) + 5(-24) = -48 - 120 = -168\]

Step 5: Solve for x, y, z Using Cramer's Rule

According to Cramer's Rule, \( x = \frac{\det(A_x)}{\det(A)} \), \( y = \frac{\det(A_y)}{\det(A)} \), \( z = \frac{\det(A_z)}{\det(A)} \).\[x = \frac{20}{50} = \frac{2}{5}\]\[y = \frac{48}{50} = \frac{24}{25}\]\[z = \frac{-168}{50} = -\frac{84}{25}\]

Key Concepts

system of equations

A system of equations consists of multiple equations with several unknown variables related to each other. Solving a system of equations means finding a set of values for these variables that satisfy all the equations. In essence, you are finding the exact point, if any, where all the equations intersect.

For example, you can imagine each equation as a line on a graph. Where lines intersect represents the solution. In our exercise, we have three equations like these:

• Equation 1: \(2x - 5y = 4\)
• Equation 2: \(x + y - z = 8\)
• Equation 3: \(3x + 5z = 0\)

These equations share three unknowns: \(x\), \(y\), and \(z\). Solving this, using techniques like substitution, elimination, or Cramer's Rule, reveals the specific values for these variables that satisfy all equations simultaneously.

determinant

The determinant is a special value that you can calculate from a square matrix. It provides important properties about the matrix, such as whether the matrix has an inverse or not. A non-zero determinant indicates that the matrix is invertible and thus has a unique solution in a system of equations.

When using Cramer's Rule, the determinant plays a crucial role in determining if a unique solution exists. For instance, if the determinant of the coefficient matrix is zero, the system likely doesn't have a unique solution.

In your given exercise, we calculate the determinant of the 3x3 matrix \( A \) as:

• \( \det(A) = 2 \cdot (5) + 5 \cdot 8 = 50 \).

This non-zero value indicates a unique solution can be found using Cramer's Rule.

matrix algebra

Matrix algebra involves operations on matrices, such as addition, multiplication, and finding determinants. It provides tools to solve systems of equations efficiently, especially when dealing with multiple variables.

In the context of solving equations using Cramer's Rule, matrix algebra simplifies the process by translating the problem into operations on arrays of numbers, making computations straightforward and organized.

During the exercise resolution, we form specific matrices from our original coefficient matrix by replacing columns one at a time with the constants from our equations. These matrices, \(A_x, A_y,\) and \(A_z\), help in finding each variable’s value when applying Cramer's Rule. This illustrates how matrix operations are crucial in solving complex systems.

linear algebra

Linear algebra is the branch of mathematics concerning linear equations, matrices, vectors, and their transformations. It provides the theoretical background to systems of equations, determinants, and matrix operations, ensuring a deeper understanding of how these processes work.

In real-world applications, linear algebra is everywhere - from computer graphics to machine learning. It's foundational in describing complexities in a simplified mathematical form.

For instance, creating the matrix from the system of equations and utilizing determinants are practices in linear algebra. Using Cramer's Rule is a direct application showcasing linear algebra's power to provide solutions in a structured manner, teaching us efficiency and precision in dealing with mathematical problems.