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91Ó°ÊÓ

Use matrices to solve the system of equations, if possible. Use Gaussian elimination with back-substitution. $$\left\\{\begin{aligned} x+2 y &=7 \\ 2 x+y &=8 \end{aligned}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = 3\) and \(y = 2\).

Step by step solution

01

Form the Augmented Matrix

An augmented matrix is a way to represent a system of equations in the form of a matrix. The given system becomes the augmented matrix: \[ \begin{bmatrix} 1 & 2 & 7 \\ 2 & 1 & 8 \end{bmatrix} \]. The vertical line separates the coefficients from the constants.
02

Apply Gaussian Elimination

Exchange rows one and two to bring the largest absolute value of a pivot to the top, and replace the second row by the difference between the double of the first row and the second row. The aim here is to get zeros below the leading coefficient (pivot) of the first row: \[ \begin{bmatrix} 2 & 1 & 8 \\ 0 & 3 & 6 \end{bmatrix} \].
03

Back-substitution to get the solution

From the second row of the final matrix obtained in step 2, it can be seen that \(3y = 6\), which simplifies to \(y = 2\). Substituting \(y = 2\) in the top row of the matrix, we get \(2x + 1*2 = 8\), which simplifies to \(2x = 6\) and finally to \(x = 3\).

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Key Concepts

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

Augmented Matrix
An augmented matrix is a compact way to write a system of linear equations. This matrix is split into two sections: the left side represents the coefficients of the variables, and the right side represents the constants from the equations' right-hand side. For example, considering a system of equations like:
\begin{align*} x + 2y &= 7; \ 2x + y &= 8, dddddddyou create the augmented matrix by placing the coefficients and the constants into a matrix form:ewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlinedddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineddddddddddd\(}\[\](\begin{bmatrix} 1 & 2 & \vert;& 7 \ 2 & 1 & \vert; & 8 \end{bmatrix})\), where the vertical line represents the separation between coefficients and constants. Students must remember that arranging the coefficients of variables and constants into the augmented matrix is a crucial initial step in applying methods like Gaussian elimination.
Back-substitution
Once a matrix has been transformed into an upper triangular form through Gaussian elimination, back-substitution is used to find the solutions to the system of equations. This method starts at the bottom row of the matrix, where you can solve for one variable directly. Then, you use this information to substitute back into the previous rows, working your way upward, to solve for the remaining variables one at a time. This process continues until all variables are solved. With the matrix \[(\begin{bmatrix} 2 & 1 & \vert; & 8 \ 0 & 3 & \vert; & 6 \end{bmatrix})\], you can see we start with the second row to find that \(y = 2\), then use that value to find \(x = 3\) from the first row. Back-substitution requires careful arithmetic to avoid mistakes and ensure accurate solutions.
Systems of Equations
A system of equations is a set of two or more equations that share a set of unknowns and are solved simultaneously. The goal is to find a set of values for the variables that satisfies all the equations at the same time. Solutions can be discovered graphically by finding the point where the equation graphs intersect, or algebraically using methods like substitution, elimination, or matrix operations like Gaussian elimination. In cases where the system does not have a unique solution, it may be determined to be dependent, having an infinite number of solutions, or inconsistent, having no solution at all.
Pivot in Linear Algebra
In linear algebra, a pivot refers to the first non-zero element in a row of a matrix, going from top to bottom and left to right. During Gaussian elimination, it's critical to choose an appropriate pivot to clear out nonzero entries below it and create an upper triangular matrix where back-substitution can commence. It is sometimes necessary to switch rows or columns in the matrix to bring a larger (by absolute value) pivot to the top for numerical stability. This helps minimize round-off errors and provides a clearer path towards solving the system of equations.
Matrix Representation
Matrices represent systems of equations, transformations in space, and more, making them a powerful tool in linear algebra. In the context of systems of equations, each row of a matrix corresponds to an equation, and each column corresponds to a coefficient in front of a variable. An augmented matrix is a special form of matrix representation that includes the constants of the equations. This concise representation allows efficient manipulation using matrix operations, which can lead to finding solutions to the system of equations. Understanding matrix representation and operations is vital for fields such as engineering, physics, and computer science, where linear algebra is heavily applied.

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Most popular questions from this chapter

Write the matrix in row-echelon form. Remember that the row-echelon form of a matrix is not unique. $$\left[\begin{array}{rrrr} 1 & -3 & 0 & -7 \\ -3 & 10 & 1 & 23 \\ 4 & -10 & 2 & -24 \end{array}\right]$$

Solve for \(x\) $$\left|\begin{array}{rrr} 1 & x & -2 \\ 1 & 3 & 3 \\ 0 & 2 & -2 \end{array}\right|=0$$

An augmented matrix that represents a system of linear equations (in the variables \(x\) and \(y\) or \(x, y,\) and \(z\) ) has been reduced using Gauss-Jordan elimination. Write the solution represented by the augmented matrix. $$\left[\begin{array}{lllll} 1 & 0 & 0 & \vdots & -4 \\ 0 & 1 & 0 & \vdots & -8 \\ 0 & 0 & 1 & \vdots & 2 \end{array}\right]$$

Write the system of linear equations represented by the augmented matrix. Then use back-substitution to find the solution. (Use the variables \(x, y,\) and \(z,\) if applicable.) $$\left[\begin{array}{rrrrr} 1 & -1 & 4 & \vdots & 0 \\ 0 & 1 & -1 & \vdots & 2 \\ 0 & 0 & 1 & \vdots & -2 \end{array}\right]$$

Four test plots were used to explore the relationship between wheat yield \(y\) (in bushels per acre) and amount of fertilizer applied \(x\) (in hundreds of pounds per acre). The results are given by the ordered pairs \((1.0,32),(1.5,41),(2.0,48),\) and (2.5,53) (a) Find the least squares regression line \(y=a x+b\) for the data by solving the system for \(a\) and \(b\) \(\left\\{\begin{array}{l}4 b+7.0 a=174 \\ 7 b+13.5 a=322\end{array}\right.\) (b) Use the regression feature of a graphing utility to confirm the result in part (a). (c) Use the graphing utility to plot the data and graph the linear model from part (a) in the same viewing window. (d) Use the linear model from part (a) to predict the yield for a fertilizer application of 160 pounds per acre.

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