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Use Cramer's rule to solve each system of equations, if possible. $$\begin{aligned} &7 x-3 y=-29\\\ &5 x+2 y=0 \end{aligned}$$

Short Answer

Expert verified
The solution is \(x = -2\) and \(y = 5\).

Step by step solution

01

Identify the coefficients matrix

First, identify the coefficients of the variables from the given system of equations. For the equations \(7x - 3y = -29\) and \(5x + 2y = 0\), the coefficients matrix \(A\) is:\[A = \begin{bmatrix} 7 & -3 \ 5 & 2 \end{bmatrix}\]
02

Compute the determinant of the coefficients matrix

Next, calculate the determinant of the matrix \(A\). The determinant of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by \(ad - bc\). For this matrix:\[\text{det}(A) = (7)(2) - (-3)(5) = 14 + 15 = 29\]
03

Formulate matrices for numerators in Cramer's Rule

For Cramer's Rule, create matrices \(A_x\) and \(A_y\) by replacing the columns of \(A\) with the column of constants, \([-29, 0]^T\).For \(A_x\):\[A_x = \begin{bmatrix} -29 & -3 \ 0 & 2 \end{bmatrix}\]For \(A_y\):\[A_y = \begin{bmatrix} 7 & -29 \ 5 & 0 \end{bmatrix}\]
04

Calculate determinants for matrices A_x and A_y

Calculate the determinants of \(A_x\) and \(A_y\):\(\text{det}(A_x):\)\[\text{det}(A_x) = (-29)(2) - (0)(-3) = -58\]\(\text{det}(A_y):\)\[\text{det}(A_y) = (7)(0) - (5)(-29) = 145\]
05

Apply Cramer's Rule to find x and y

According to Cramer's Rule, solve for \(x\) and \(y\) using the formulas:\[x = \frac{\text{det}(A_x)}{\text{det}(A)}\]\[y = \frac{\text{det}(A_y)}{\text{det}(A)}\]Substitute the determinants:\[x = \frac{-58}{29} = -2\]\[y = \frac{145}{29} = 5\]
06

Conclude the solution

Therefore, the solution to the system of equations is:\[x = -2\]\[y = 5\]

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Key Concepts

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

Systems of Equations
A system of equations consists of multiple equations that share the same set of variables. In our example, we have the equations \(7x - 3y = -29\) and \(5x + 2y = 0\) with two unknowns \(x\) and \(y\). The goal is to find values that satisfy all equations in the system at once.

These types of problems can represent real-world situations where multiple conditions need to be met simultaneously. Solving systems of equations through methods like Cramer's Rule is essential in various fields, including engineering, economics, and physics.

Students need to understand that systems of equations can be tackled using several techniques, such as substitution, elimination, and matrix methods like Cramer's Rule. Each approach has its own advantage depending on the specific problem structure.
Determinant
The determinant is a unique number that is calculated from a square matrix. For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is \(ad - bc\). This determinant plays a crucial role when using Cramer's Rule to solve a system of equations.

In our exercise, the determinant of the coefficients matrix \(A\) is crucial, as it verifies if a unique solution exists. If the determinant is zero, Cramer's Rule cannot be applied, indicating that the system might be dependent or inconsistent. In contrast, a non-zero determinant, like the 29 we computed, ensures a unique solution for the system.
  • Determinant tells us about the matrix invertibility.
  • Determinant helps us understand volume scaling when a matrix is seen as a transformation.
Matrices
A matrix is a rectangular array of numbers, which is a central tool for organizing and solving systems of equations. In our example, we used a 2x2 matrix \(A\) to organize the coefficients from the system of linear equations.

By representing equations in a matrix form, it becomes easier to manipulate and solve them using various algebraic methods. Cramer's Rule, which is based on determinants and matrices, provides a neat way to find solutions to these systems.

In this exercise, matrices \(A_x\) and \(A_y\) were formed by replacing different columns of \(A\) with the constants from the system. This step is critical for using Cramer's Rule, where the focus shifts from the original matrix to these newly formed matrices for solution calculations.
Precalculus
Precalculus is a mathematical course that serves as a bridge between algebra, trigonometry, and calculus. It often includes the study of mathematical tools like systems of equations, matrices, and determinants. These concepts are foundational for the more advanced topics in calculus.

Learning about matrix operations and methods for solving systems of equations, such as Cramer's Rule, sets a strong base for calculus. Understanding Cramer's Rule involves both algebraic manipulation and an appreciation of the determinant concept, blending linear algebra techniques within the precalculus scope.

These precalculus concepts are essential for solving practical problems and preparing for calculus challenges. By mastering these topics, students enhance their problem-solving skills and readiness for more complex mathematical ideas.

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Most popular questions from this chapter

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. $$\begin{array}{|c|c|} \hline t \text { (seconos) } & \text { HeiGHT (FEET) } \\ \hline 1 & 34 \\ \hline 2 & 36 \\ \hline 3 & 6 \\ \hline \end{array}$$

In order for \(A_{m \times n}^{2}\) to be defined, what condition (with respect to \(m\) and \(n\) ) must be met?

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

In calculus, determinants are used when evaluating double and triple integrals through a change of variables. In these cases, the elements of the determinant are functions. Find each determinant. $$\left|\begin{array}{cc} 2 x & 2 y \\ 2 x & 2 y-2 \end{array}\right|$$

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?

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