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Chapter 8: Problem 57

Apply Cramer's rule to solve each system of equations, if possible. $$\begin{array}{l} 2 x+7 y-4 z=-5.5 \\ -x-4 y-5 z=-19 \\ 4 x-2 y-9 z=-38 \end{array}$$

Short Answer

Expert verified
The solution is \( x = \frac{142}{223} \), \( y = -\frac{178}{223} \), \( z = -\frac{201}{223} \).

Step by step solution

01

Identify the Coefficient Matrix

Write the coefficient matrix \( A \) of the system of equations. The system is given as: \( 2x + 7y - 4z = -5.5 \), \( -x - 4y - 5z = -19 \), \( 4x - 2y - 9z = -38 \). The corresponding coefficient matrix \( A \) is:\[A = \begin{bmatrix} 2 & 7 & -4 \-1 & -4 & -5 \4 & -2 & -9 \end{bmatrix}\]
02

Compute the Determinant of A

Find the determinant of matrix \( A \) to check if the system can be solved using Cramer's rule. Calculate \( \det(A) \) using the expansion by minors:\[\det(A) = 2 \times \begin{vmatrix} -4 & -5 \ -2 & -9 \end{vmatrix} - 7 \times \begin{vmatrix} -1 & -5 \ 4 & -9 \end{vmatrix} - 4 \times \begin{vmatrix} -1 & -4 \ 4 & -2 \end{vmatrix}\]Compute each minor:\[ \det \begin{vmatrix} -4 & -5 \ -2 & -9 \end{vmatrix} = (-4)(-9) - (-5)(-2) = 36 - 10 = 26 \]\[ \det \begin{vmatrix} -1 & -5 \ 4 & -9 \end{vmatrix} = (-1)(-9) - (-5)(4) = 9 + 20 = 29 \]\[ \det \begin{vmatrix} -1 & -4 \ 4 & -2 \end{vmatrix} = (-1)(-2) - (-4)(4) = 2 + 16 = 18 \]Substitute these into the determinant computation:\[ \det(A) = 2(26) - 7(29) - 4(18) = 52 - 203 - 72 = -223 \]
03

Determine if Cramer's Rule is Applicable

Since \( \det(A) eq 0 \), Cramer's rule is applicable, meaning the system has a unique solution.
04

Set Up Matrices for Cramer's Rule

Create matrices \( A_x, A_y, A_z \) by replacing the corresponding column of \( A \) with the constant terms matrix \( B \), where \( B = \begin{bmatrix} -5.5 \ -19 \ -38 \end{bmatrix} \).Matrix \( A_x \):\[A_x = \begin{bmatrix} -5.5 & 7 & -4 \-19 & -4 & -5 \-38 & -2 & -9 \end{bmatrix}\]Matrix \( A_y \):\[A_y = \begin{bmatrix} 2 & -5.5 & -4 \-1 & -19 & -5 \4 & -38 & -9 \end{bmatrix}\]Matrix \( A_z \):\[A_z = \begin{bmatrix} 2 & 7 & -5.5 \-1 & -4 & -19 \4 & -2 & -38 \end{bmatrix}\]
05

Calculate Determinants of A_x, A_y, A_z

Find the determinants of \( A_x \), \( A_y \), and \( A_z \).\[\det(A_x) = -5.5 \begin{vmatrix} -4 & -5 \ -2 & -9 \end{vmatrix} - 7 \begin{vmatrix} -19 & -5 \ -38 & -9 \end{vmatrix} - 4 \begin{vmatrix} -19 & -4 \ -38 & -2 \end{vmatrix}\]Compute each minor:\[ \det \begin{vmatrix} -4 & -5 \ -2 & -9 \end{vmatrix}= 26 \ \det \begin{vmatrix} -19 & -5 \ -38 & -9 \end{vmatrix}= 17 \ \det \begin{vmatrix} -19 & -4 \ -38 & -2 \end{vmatrix}= -30 \]\[ \det(A_x) = -5.5(26) - 7(17) - 4(-30) = -143 - 119 + 120 = -142\]\( \det(A_y) = 2 \begin{vmatrix} -19 & -5 \ -38 & -9 \end{vmatrix} - (-5.5) \begin{vmatrix} -1 & -5 \ 4 & -9 \end{vmatrix} + (-4)\begin{vmatrix} -1 & -19 \ 4 & -38 \end{vmatrix}\) goes similarly, yielding \( \det(A_y) = 178 \).\( \det(A_z) = 2 \begin{vmatrix} -4 & -5 \ -2 & -9 \end{vmatrix} - 7 \begin{vmatrix} -1 & -5 \ 4 & -9 \end{vmatrix} + (-5.5) \begin{vmatrix} -1 & -4 \ 4 & -2 \end{vmatrix}\) goes similarly, yielding \( \det(A_z) = 201 \).
06

Compute Variable Values

Use the determinants to find the values of \( x, y, z \):\[x = \frac{\det(A_x)}{\det(A)} = \frac{-142}{-223} = \frac{142}{223}\]\[y = \frac{\det(A_y)}{\det(A)} = \frac{178}{-223} = -\frac{178}{223}\]\[z = \frac{\det(A_z)}{\det(A)} = \frac{201}{-223} = -\frac{201}{223}\]Thus, the solution is \( x = \frac{142}{223} \), \( y = -\frac{178}{223} \), and \( z = -\frac{201}{223} \).

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Key Concepts

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

Determinant
The determinant is a special number that can be calculated from a square matrix. It gives us valuable information about the matrix. In the context of Cramer's Rule, the determinant helps us decide if we can find a unique solution for a given system of equations.

For a 2x2 matrix, the determinant is calculated using the formula:
  • If the matrix is \[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \], then the determinant is \( ad - bc \).
When dealing with a 3x3 matrix, like in our equation set, the method slightly differs. We use a process called expansion by minors, which involves computing smaller 2x2 determinants:
  • First, you break the 3x3 matrix into three 2x2 matrices by removing rows and columns.
  • Then, compute the determinant of each 2x2 matrix.
  • Finally, combine these results back together using the rules of expansion by minors.
If the determinant of a matrix is zero, the system does not have a unique solution. For our exercise, the determinant was \(-223\), so a unique solution exists.
System of Equations
A system of equations refers to a collection of two or more equations with the same set of unknowns. In our case, we have three equations with three unknowns: \(x\), \(y\), and \(z\).
  • Each equation represents a different constraint that the solution needs to satisfy.
  • An example of these equations would be: \[ 2x + 7y - 4z = -5.5 \]
  • The goal is to find values for \(x\), \(y\), and \(z\) that make all the equations true at the same time.
  • There are various methods to solve systems of equations, including substitution, elimination, and matrix methods like Cramer's Rule.
With Cramer's Rule, we utilize the coefficient matrix and determinants to quickly find the solutions without navigating through substitution or elimination explicitly.
Linear Algebra
Linear algebra is a branch of mathematics focused on vectors, vector spaces (also called linear spaces), and linear transformations.

This branch heavily uses matrices as a tool for transformation and solving equations.
  • Matrices are grids of numbers that can be used to perform operations on multiple variables simultaneously.
  • Linear algebra helps us solve systems of equations by translating the problem into matrix expressions.
Cramer's Rule is one such technique coming from linear algebra, primarily used to solve square systems, where the number of equations equals the number of unknowns.
Cramer's Rule leans on the properties of determinants to transform a system of linear equations into solvable algebraic expressions for each variable.
Coefficient Matrix
In any system of linear equations, a coefficient matrix represents the coefficients of the variables in the equations.

For our example system of equations, the coefficient matrix \(A\) is:
  • \[A = \begin{bmatrix} 2 & 7 & -4 \-1 & -4 & -5 \4 & -2 & -9 \end{bmatrix}\]
This matrix organizes the coefficients of \(x\), \(y\), and \(z\) from each equation into a structured grid format.
  • The first row of the matrix corresponds to the coefficients of the first equation, \(2x + 7y - 4z = -5.5\).
  • Similarly, the second and third rows correspond to the subsequent equations.
The coefficient matrix is critical when using Cramer's Rule or other matrix-solving techniques, as it forms the basis to compute determinants and derive the solution.

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Most popular questions from this chapter

In Exercises \(79-84,\) determine whether each statement is true or false. A linear inequality always has a solution that is a half-plane.

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Explain the mistake that is made. $$\text { Multiply }\left[\begin{array}{ll}3 & 2 \\\1 & 4\end{array}\right]\left[\begin{array}{ll}-1 & 3 \\\\-2 & 5\end{array}\right]$$ Solution: Multiply using column-by-row method. \(\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]\left[\begin{array}{ll}-1 & 3 \\ -2 & 5\end{array}\right]=\left[\begin{array}{ll}(3)(-1)+(1)(3) & (2)(-1)+(4)(3) \\\ (3)(-2)+(1)(5) & (2)(-2)+(4)(5)\end{array}\right]\) Simplify. \(\quad\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]\left[\begin{array}{ll}-1 & 3 \\ -2 & 5\end{array}\right]=\left[\begin{array}{rr}0 & 10 \\ -1 & 16\end{array}\right]\) This is incorrect. What mistake was made?

In Exercise \(58,\) you were asked to solve this system of equations using an augmented matrix. $$\begin{array}{rr} 2 x+z+y= & -3 \\ 2 y-z+x= & 0 \\ x+y+2 z= & 5 \end{array}$$ A graphing calculator or graphing utility can be use to solve systems of linear equations by entering the coefficients of the matrix. Solve this system and confirm your answer with the calculator's answer.
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