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Use Cramer's Rule to solve the system of equations. $$\left\\{\begin{array}{rr} 7 x-y= & -8 \\ -x+3 y= & 4 \end{array}\right.$$

Short Answer

Expert verified
The solution to the system of equations is x = -1 and y = 1.

Step by step solution

01

Compute Main Determinant (D)

The determinant will be computed utilizing the coefficients of x and y. Here, \( D = \begin{vmatrix} 7 & -1 \ -1 & 3 \end{vmatrix} = 7*3 - (-1)*(-1) = 20.
02

Compute X Determinant (Dx)

Retain the coefficients of y and swap coefficients of x with the solution values to compute Dx. \( Dx = \begin{vmatrix} -8 & -1 \ 4 & 3 \end{vmatrix} = -24 - (-4) = -20.
03

Compute Y Determinant (Dy)

Retain the coefficients of x and swap coefficients of y with the solution values to compute Dy. \( Dy = \begin{vmatrix} 7 & -8 \ -1 & 4 \end{vmatrix} = 28 - 8 = 20.
04

Compute X and Y

The values of x and y can be found by dividing Dx by D and Dy by D, respectively. \( x = Dx/D = -20/20 = -1 \) and \( y = Dy/D = 20/20 = 1 \).

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Key Concepts

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

System of Equations
In algebra, a system of equations is a set of multiple equations, each with the same variables, which all need to be satisfied simultaneously. The solution to a system of equations is the set of variable values that make all equations true at the same time. For example, the combined solution of equations such as \(7x - y = -8\) and \(-x + 3y = 4\) requires finding values for \(x\) and \(y\) that satisfy both equations.

To solve a system like this, various methods can be used, including graphing, substitution, elimination, and matrix-based methods. Graphing involves plotting each equation on a graph to find the point where they intersect, which represents the solution. Substitution and elimination are algebraic techniques where you solve for one variable in terms of the other, or add/subtract equations to eliminate a variable. It's important for students to understand the concept behind these methods to apply them effectively to different types of problems.
Determinants
The determinant is a special number that can be calculated from a square matrix. In linear algebra, the determinant helps us find out if a system of linear equations has a unique solution, no solution, or infinitely many solutions. It is also used in calculating the inverse of a matrix and in describing geometric properties of linear transformations.

To calculate a 2x2 matrix determinant, for the matrix \(\begin{vmatrix} a & b \ c & d \/\begin{vmatrix}\), the determinant would be \(ad - bc\). When applying Cramer's Rule to solve a system of equations, determinants become crucial as they aid in finding the solution for the variables. If the main determinant (D) is non-zero, the system is consistent and has a unique solution, just like in our example with \(D = 7*3 - (-1)*(-1) = 20\). However, if the determinant is zero, the system may have no solution or infinitely many solutions, and Cramer's Rule cannot be applied.
Algebraic Solutions
Finding algebraic solutions to systems of equations involves steps that manipulate the equations algebraically to isolate the variables. Cramer's Rule is one algebraic method that provides a straightforward process for systems that have as many equations as unknowns, and where the determinant of the coefficients of the variables is not zero. It makes use of determinants to solve each variable individually.

The process, as shown in the solution to our exercise, includes calculating the determinant for each variable's matrix (Dx for x and Dy for y) by replacing the respective column of the coefficients with the constants from the right-hand side of the equations. The variable is found by dividing each determinant by the main determinant (D). This results in values \(x = -1\) and \(y = 1\), which are the algebraic solutions to the system. It's important to grasp the beauty of algebra as it turns complex systems into solvable equations with clear answers.

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Most popular questions from this chapter

Find \(\left(A^{2}\right)^{-1}\) and \(\left(A^{-1}\right)^{2},\) where \(A=\left[\begin{array}{rr}1 & -2 \\ -1 & 3\end{array}\right] .\) What do you observe? Use the definition of the inverse of a matrix, together with the fact that \((A B)^{-1}=A^{-1} B^{-1},\) to show that \(\left(A^{2}\right)^{-1}=\left(A^{-1}\right)^{2}\) for every square matrix \(A\)

A couple has \(\$ 10,000\) to invest for their child's wedding. Their accountant recommends placing at least \(\$ 6000\) in a high-yield investment and no more than \(\$ 4000\) in a low-yield investment. (a) Use \(x\) to denote the amount of money placed into the high-yield investment. Use \(y\) to denote the amount of money placed into the low-yield investment. Write a system of linear inequalities that describes the possible amounts the couple could invest in each type of venture. (b) Graph the region that represents all possible amounts the couple could put into each investment if they wish to follow the accountant's advice.

Answer the question pertaining to the matrices. $$A=\left[\begin{array}{ll}a & b \\\c & d \\\e & f\end{array}\right] \text { and } B=\left[\begin{array}{lll}g & h & i \\\j & k & l\end{array}\right]$$ Let \(Q=B A,\) and find \(q_{11}\) and \(q_{22}\) without performing the entire multiplication of matrix \(B\) by matrix \(A\).

For the given matrices \(A, B,\) and \(C,\) evaluate the indicated expression. $$\begin{aligned}&A=\left[\begin{array}{ll}4 & 1 \\\0 & 2 \\\5 & 1\end{array}\right] ; \quad B=\left[\begin{array}{rr}4 & 3 \\\\-6 & 2 \\\3 & -1\end{array}\right]\\\&C=\left[\begin{array}{rrr}1 & 2 & 3 \\\\-2 & -3 & -1 \\\3 & 1 & 2\end{array}\right] ; \quad C(B-A)\end{aligned}$$

The sum of money invested in two savings accounts is \(\$ 1000 .\) If both accounts pay \(4 \%\) interest compounded annually, is it possible to earn a total of \(\$ 50\) in interest in the first year? (a) Explain your answer in words. (b) Explain your answer using a system of equations.

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