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Chapter 8: Problem 12

Use Cramer's rule to solve each system of equations, if possible. $$\begin{array}{l} x+y=-1 \\ x-y=-9 \end{array}$$

Short Answer

Expert verified
The solution is \(x = -5\) and \(y = 4\).

Step by step solution

01

Determine Matrix Representation

First, initialize the system of equations in matrix form: \[\begin{cases}x + y = -1 \x - y = -9\end{cases}\] can be written as \(AX = B\), where \(A = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix}\), \(X = \begin{pmatrix} x \ y \end{pmatrix}\), and \(B = \begin{pmatrix} -1 \ -9 \end{pmatrix}\).
02

Calculate Determinant of Matrix A

Calculate the determinant of matrix \(A\). For a \(2 \times 2\) matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is \(ad - bc\). So for \[A = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix}, \\text{Det}(A) = (1)(-1) - (1)(1) = -1 - 1 = -2\] The determinant is \(-2\), which is non-zero, so the system has a unique solution.
03

Determine Determinant of Modified Matrices

Calculate the determinants of the matrices \(A_x\) and \(A_y\) by replacing the respective columns of \(A\) with \(B\). For \(A_x\): Replace the first column of \(A\) with \(B\):\[A_x = \begin{pmatrix} -1 & 1 \ -9 & -1 \end{pmatrix}\]Compute:\[\text{Det}(A_x) = (-1)(-1) - (1)(-9) = 1 + 9 = 10\]For \(A_y\): Replace the second column of \(A\) with \(B\):\[A_y = \begin{pmatrix} 1 & -1 \ 1 & -9 \end{pmatrix}\]Compute:\[\text{Det}(A_y) = (1)(-9) - (-1)(1) = -9 + 1 = -8\]
04

Use Cramer's Rule to Find Solutions

Use Cramer's rule to solve for \(x\) and \(y\). The rule states: \[x = \frac{\text{Det}(A_x)}{\text{Det}(A)}\] \[y = \frac{\text{Det}(A_y)}{\text{Det}(A)}\]Calculate:\[x = \frac{10}{-2} = -5\]\[y = \frac{-8}{-2} = 4\]Therefore, the solution is \(x = -5\) and \(y = 4\).

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Key Concepts

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

Matrix Representation
When dealing with a system of equations, representing them in matrix form can streamline the process of finding solutions. A system like:
  • \[x + y = -1\]
  • \[x - y = -9\]
can be transformed into a matrix equation \(AX = B\). Here:
  • \(A\) is the coefficient matrix: \(\begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix}\).
  • \(X\) is the column matrix of variables: \(\begin{pmatrix} x \ y \end{pmatrix}\).
  • \(B\) is the constant matrix: \(\begin{pmatrix} -1 \ -9 \end{pmatrix}\).
This transformation helps in applying mathematical rules, such as Cramer's Rule, to find the solution efficiently and accurately.
Understanding and setting up these matrices correctly is crucial for the next steps.
Determinant Calculation
The next step in using Cramer's Rule is calculating the determinant of the coefficient matrix \(A\). For any \(2 \times 2\) matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is calculated as \(ad - bc\). In our case:
  • \(A = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix}\)
  • Determinant of \(A\), \(\text{Det}(A) = (1)(-1) - (1)(1) = -1 - 1 = -2\)
The determinant is non-zero, which is significant. A non-zero determinant indicates that the system has a unique solution. If the determinant were zero, it would suggest that the system may have either no solution or an infinite number of solutions. Hence, this calculation is pivotal to determine if the system can be uniquely solved.
System of Equations
A system of equations involves multiple equations that are solved together. In this case, the system is:
  • \(x + y = -1\)
  • \(x - y = -9\)
In real-world problems, systems of equations can arise in various contexts such as physics for solving forces, economics for determining market equilibria, or everyday situations needing more than one relationship to describe the problem.
Solving a system can be approached in many ways, but Cramer's Rule is particularly useful when the system is linear and can be represented by matrices. If the system's coefficient matrix has a non-zero determinant, Cramer's Rule provides a straightforward method utilizing determinants to find the unique values for the variables involved.
Unique Solution
The concept of a unique solution is central to solving systems of equations using Cramer's Rule. In our example, we found:
  • \(x = \frac{10}{-2} = -5\)
  • \(y = \frac{-8}{-2} = 4\)
This means that the values \(x = -5\) and \(y = 4\) satisfy both equations in the system.
A unique solution exists when every variable has a specific value that satisfies all equations simultaneously. The presence of a non-zero determinant guarantees that for a set of linear equations, such a unique solution can be found.
This is crucial for ensuring that the solution is not just mathematically viable but also consistent with potential real-world scenarios or any constraints of the problem itself.

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Most popular questions from this chapter

Use the following tables. The following table gives fuel and electric requirements per mile associated with gasoline and electric automobiles: $$\begin{array}{|l|c|c|}\hline & \begin{array}{c} \text { Numeer of } \\\\\text { GatLons/Mile }\end{array} & \begin{array}{c}\text { Numeer of } \\\\\text { kW-hr/Mile }\end{array} \\\\\hline \text { SUV full size } & 0.06 & 0 \\ \hline \text { Hybrid car } & 0.02 & 0.1 \\\\\hline \text { Electric car } & 0 & 0.3 \\\\\hline\end{array}$$ The following table gives an average cost for gasoline and electricity: $$\begin{array}{|l|c|}\hline \text { Cost per gallon of gasoline } & \$ 3.80 \\\\\hline \text { Cost per kW-hr of electricity } & \$ 0.05 \\\\\hline\end{array}$$ Environment. Let matrix \(A\) represent the gasoline and electricity consumption and matrix \(B\) represent the costs of gasoline and electricity. Find \(A B\) and describe what the elements of the product matrix represent. (Hint: \(A\) has order \(3 \times 2\) and \(B\) has order \(2 \times 1 .)\)

Explain the mistake that is made. Perform the indicated row operations on the matrix. \(\left[\begin{array}{cccc}1 & -1 & 1 & 2 \\ 2 & -3 & 1 & 4 \\ 3 & 1 & 2 & -6\end{array}\right]\) a. \(R_{2}-2 R_{1} \rightarrow R_{2}\) b. \(R_{3}-3 R_{1} \rightarrow R_{3}\) Solution: a. \(\left[\begin{array}{rrr|r}1 & -1 & 1 & 2 \\ 0 & -3 & 1 & 4 \\ 3 & 1 & 2 & -6\end{array}\right]\) b. \(\left[\begin{array}{rrr|r}1 & -1 & 1 & 2 \\ 2 & -3 & 1 & 4 \\ 0 & 1 & 2 & -6\end{array}\right]\) This is incorrect. What mistake was made?

Apply a graphing utility to perform the indicated matrix operations, if possible. $$A=\left[\begin{array}{rrrr}1 & 7 & 9 & 2 \\\\-3 & -6 & 15 & 11 \\\0 & 3 & 2 & 5 \\\9 & 8 & -4 & 1\end{array}\right] \quad B=\left[\begin{array}{rr}7 & 9 \\\8 & 6 \\\\-4 & -2 \\\3 & 1\end{array}\right]$$ $$A B$$

Solve the system of linear equations. $$ \begin{array}{l} 4 x-6 y=0 \\ 4 x+6 y=4 \end{array} $$ Solution: Set up the determinants. $$ D=\left|\begin{array}{rr} 4 & -6 \\ 4 & 6 \end{array}\right|, D_{x}=\left|\begin{array}{rr} 0 & -6 \\ 4 & 6 \end{array}\right|, \text { and } D_{y}=\left|\begin{array}{ll} 4 & 0 \\ 4 & 4 \end{array}\right| $$ Evaluate the determinants. $$ \begin{array}{c} D=48, D_{x}=24, \text { and } D_{y}=16 \\ x=\frac{D}{D_{x}}=\frac{48}{24}=2 \text { and } y=\frac{D_{y}}{D}=\frac{48}{16}=3 \end{array} $$ Solve for \(x\) and \(y\) \(x=2, y=3\) is incorrect. What mistake was made?

Involve vertical motion and the effect of gravity on an object. Because of gravity, an object that is projected upward will eventually reach a maximum height and then fall to the ground. The equation that relates the height \(h\) of a projectile \(t\) seconds after it is projected upward is given by $$h=\frac{1}{2} a t^{2}+v_{0} t+h_{0}$$ where \(a\) is the acceleration due to gravity, \(h_{0}\) is the initial height of the object at time \(t=0,\) and \(v_{0}\) is the initial velocity of the object at time \(t=0 .\) Note that a projectile follows the path of a parabola opening down, so \(a<0\). An object is thrown upward, and the table below depicts the height of the ball \(t\) seconds after the projectile is released. Find the initial height, initial velocity, and acceleration due to gravity. $$\begin{array}{|c|c|} \hline t \text { (seconos) } & \text { Heiant (FEET) } \\ \hline 1 & 54 \\ \hline 2 & 66 \\ \hline 3 & 46 \\ \hline \end{array}$$
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