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Chapter 4: Problem 23

Use Cramer鈥檚 Rule to solve each system of equations. \(2 r+25 s=40\) \(10 r+12 s+6 t=-2\) \(36 r-25 s+50 t=-10\)

Short Answer

Expert verified
\( r = -6.32 \), \( s = 2.95 \), \( t = 0.50 \).

Step by step solution

01

Write the System of Equations

Identify the given system of equations from the problem: \(2r + 25s = 40\), \(10r + 12s + 6t = -2\), \(36r - 25s + 50t = -10\). Make sure all equations are in the form of \(Ax + By + Cz = D\).
02

Arrange Coefficients in a Matrix

Set up the coefficient matrix \(A\) from the system, which is: \[ A = \begin{bmatrix} 2 & 25 & 0 \ 10 & 12 & 6 \ 36 & -25 & 50 \end{bmatrix} \]. Also, set up the constants matrix \(B\) as \[ B = \begin{bmatrix} 40 \ -2 \ -10 \end{bmatrix} \].
03

Calculate the Determinant of Matrix A

The determinant of matrix \(A\) can be calculated using: \( \text{det}(A) = 2(12 \cdot 50 - -25 \cdot 6) - 25(10 \cdot 50 - 6 \cdot 36) + 0 = 2(600 + 150) + 6250 = 1500 - 6250 + 0 = -4750\).
04

Find Determinant for Modified Matrices

Calculate determinants of matrices where each column is replaced by \(B\).1. \(A_r\) is \[ \begin{bmatrix} 40 & 25 & 0 \ -2 & 12 & 6 \ -10 & -25 & 50 \end{bmatrix} \]. Its determinant \( \Delta_r \): \( 40(12 \cdot 50 + 25 \cdot 6) + 2 <0> - 0 \). Calculate: \( (600 + 150) = 750, 40 \cdot 750 = 30000 \).2. \(A_s\) is \[ \begin{bmatrix} 2 & 40 & 0 \ 10 & -2 & 6 \ 36 & -10 & 50 \end{bmatrix} \] and its \( \Delta_s \): \( -14000 \).3. \(A_t\) is \[ \begin{bmatrix} 2 & 25 & 40 \ 10 & 12 & -2 \ 36 & -25 & -10 \end{bmatrix} \] and its \( \Delta_t \): \( -2360 \).
05

Apply Cramer's Rule

The solutions are obtained using the formula: \( r = \frac{\Delta_r}{\text{det}(A)} \), \( s = \frac{\Delta_s}{\text{det}(A)} \), \( t = \frac{\Delta_t}{\text{det}(A)} \).\( r = \frac{30000}{-4750} \approx -6.32 \), \( s = \frac{-14000}{-4750} \approx 2.95 \), \( t = \frac{-2360}{-4750} \approx 0.50 \).
06

Final Answers

Conclude by stating the solutions: \( r = -6.32 \), \( s = 2.95 \), \( t = 0.50 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant Calculation
The calculation of the determinant is a crucial step in Cramer's Rule for solving systems of equations. To determine the value of the determinant, you begin by setting up a square matrix with the coefficients from your system of equations. In our problem, the matrix \( A \) is \( \begin{bmatrix} 2 & 25 & 0 \ 10 & 12 & 6 \ 36 & -25 & 50 \end{bmatrix} \). To find its determinant, apply the rule for 3x3 matrices:
  • Select an element from the first row; here, using \( 2 \).
  • Multiply this element by the determinant of the 2x2 matrix that remains when the row and column containing that element are deleted.
  • Repeat this process for the remaining elements in the first row, alternately subtract and add each result, adjusting for their respective positions.
For matrix \( A \), the computation yields a determinant of \(-4750\). This determinant helps determine the solvability of the system, with a non-zero value indicating a unique solution.
System of Equations
A system of equations consists of multiple equations that are solved together because they share variables. In linear algebra, these systems frequently involve equations that take the form \( Ax + By + Cz = D \). Our given equations can be simplified as follows:
  • First equation: \( 2r + 25s = 40 \)
  • Second equation: \( 10r + 12s + 6t = -2 \)
  • Third equation: \( 36r - 25s + 50t = -10 \)
The goal is to find the values of \( r \), \( s \), and \( t \) that satisfy all the equations simultaneously. This is where methods like Cramer's Rule are useful, allowing us to assess if a set of solutions exists using determinants from matrices, provided the determinant of the coefficient matrix is non-zero.
Matrix Algebra
Matrix algebra is the language of systems of equations like ours. By expressing the system in matrix form, we can use powerful tools like determinants and inverse matrices.
  • The coefficient matrix \( A \) holds the coefficients of your variables: \( \begin{bmatrix} 2 & 25 & 0 \ 10 & 12 & 6 \ 36 & -25 & 50 \end{bmatrix} \).
  • The constants matrix \( B \) (or vector) comprises the right-hand side of the equations: \( \begin{bmatrix} 40 \ -2 \ -10 \end{bmatrix} \).
Through matrix manipulation, a system of equations can be translated into a more compact form, making calculation more straightforward and systematic. In our exercise, matrix algebra simplifies the application of Cramer's Rule to derive the unknowns.
Linear Equations
Linear equations, such as those in this example, are equations of the first degree. They form a straight line when graphed on a Cartesian plane. In our given problem, they determine how quantities flow with respect to each other. Core properties include:
  • The solutions to these equations are where the graph intersects the axes, assuming each variable represents a unique dimension.
  • In a three-equation system in three unknowns, each solution set corresponds to a point where the planes defined by these equations intersect.
Linear equations often model real-world relations allowing us to predict one variable from others. Their simplicity lends itself to methods like elimination, substitution, and Cramer's Rule, helping efficiently solve for unknowns in systems by reworking them into solvable formats through algebraic manipulation.

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