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Chapter 16: Problem 4

Solve the given systems of equations by Gaussian elimination. If there is an unlimited number of solutions, find two of them. $$\begin{aligned} &2 x+y=1\\\ &5 x+2 y=1 \end{aligned}$$

Short Answer

Expert verified
The unique solution is \(x = -1\), \(y = 3\).

Step by step solution

01

Write the augmented matrix

First, translate the system of equations into an augmented matrix. For the system \(2x + y = 1\) and \(5x + 2y = 1\), the augmented matrix is:\[\begin{bmatrix}2 & 1 & | & 1 \5 & 2 & | & 1 \\end{bmatrix}\]
02

Perform row operations to get an upper triangular form

We want to eliminate the first element of the second row (5) using the first row. We do this by replacing the second row with the result of \(R_2 - \frac{5}{2}R_1\):\[R_2 = \begin{bmatrix}5 & 2 & | & 1\end{bmatrix} - \frac{5}{2}\begin{bmatrix}2 & 1 & | & 1\end{bmatrix} = \begin{bmatrix}0 & -0.5 & | & -1.5\end{bmatrix}\]The new augmented matrix is:\[\begin{bmatrix}2 & 1 & | & 1 \0 & -0.5 & | & -1.5 \\end{bmatrix}\]
03

Simplify the second row

Now simplify the second row by multiplying it by -2 to make it easier to interpret:\[R_2 = -2 \times \begin{bmatrix}0 & -0.5 & | & -1.5\end{bmatrix} = \begin{bmatrix}0 & 1 & | & 3\end{bmatrix}\]After simplification, the augmented matrix is:\[\begin{bmatrix}2 & 1 & | & 1 \0 & 1 & | & 3 \\end{bmatrix}\]
04

Back substitution to solve for variables

Using back substitution, solve for \(y\) from the second row: \(y = 3\).Substitute \(y = 3\) into the first equation: \(2x + 3 = 1\). Solve for \(x\):\[2x = 1 - 3\]\[2x = -2\]\[x = -1\]
05

Verify the solution

Substitute \(x = -1\) and \(y = 3\) back into the original equations to verify the solution.First equation: \(2(-1) + 3 = 1\) is true.Second equation: \(5(-1) + 2(3) = 1\) is true.Both equations are satisfied.

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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 two or more equations with the same set of variables. In this exercise, we have a system with two equations:
  • \(2x + y = 1\) and
  • \(5x + 2y = 1\).
The goal is to find the values of the variables that satisfy all equations simultaneously. There are different methods to solve systems of equations, such as graphing, substitution, and elimination. Gaussian elimination, the focus of this exercise, is a systematic way to simplify and solve these systems by transforming them into a more straightforward form, making it easier to find solutions.
It is important to look for both unique solutions and the possibility of multiple solutions depending on the relationship between the equations.
Augmented Matrix
An augmented matrix is a compact way of representing a system of linear equations. Each row corresponds to an equation, and each column corresponds to a variable or an equation's constant term. In our example, the augmented matrix looks like this:
  • \[\begin{bmatrix}2 & 1 & | & 1 \5 & 2 & | & 1\end{bmatrix}\].
The vertical bar separates the coefficients of the variables on the left from the constants on the right. Augmented matrices allow us to apply row operations systematically to transform the system into an easily solvable form. They're particularly useful in Gaussian elimination since they allow the equations' coefficients to be manipulated to achieve the upper triangular form.
Upper Triangular Form
The upper triangular form is key in simplifying systems of equations. It involves transforming the augmented matrix so that all elements below the main diagonal are zeros. For our exercise, this was achieved through the row operation:
  • \( R_2 = R_2 - \frac{5}{2}R_1 \),
resulting in the following matrix:
  • \[\begin{bmatrix}2 & 1 & | & 1 \0 & -0.5 & | & -1.5\end{bmatrix}\].
This form is advantageous because it allows easier application of back substitution. Simplifying the second row further, by making leading coefficients equal to 1, helps provide direct solutions for the unknowns, making the process more efficient and straightforward.
Back Substitution
Back substitution is the final step in solving the system of equations after achieving an upper triangular form. This process involves starting with the last row of the upper triangular matrix and finding the value of one variable. In our exercise, starting from the second row yielded:
  • \( y = 3 \).
Next, you substitute this value back into the first equation to find the value of \( x \):
  • \( 2x + 3 = 1 \), leading to \( x = -1 \).
Back substitution is efficient because it reduces the equations one by one, using known values to find unknowns systematically. This method works reliably once the matrix is in an upper triangular form, ensuring the solution is accurate upon verification.

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Most popular questions from this chapter

Perform the indicated matrix operations. The contractor of a housing development constructs four different types of houses, with either a carport, a one-car garage, or a two-car garage. The following matrix shows the number of houses of each type and the type of garage. $$\begin{array}{c|c|} \text { }& \text { Type A } & \text { Type B } & \text { Type C } & \text { Type D } \\ \text { Carport } & 96 & 75 & 0 & 0 \\ \text { 1-car garage } & 64 & 44 & 24 & 0 \\ \text { 2-car garage } & 0 & 35 & 68 & 78 \\ \end{array}$$ If the contractor builds two additional identical developments, find the matrix showing the total number of each house-garage type built in the three developments.

Perform the indicated matrix multiplications on a calculator, using the following matrices. For matrix \(A, A^{2}=A \times A\) $$A=\left[\begin{array}{rrr} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{array}\right] B=\left[\begin{array}{rrr} 1 & -2 & -6 \\ -3 & 2 & 9 \\ 2 & 0 & -3 \end{array}\right] C=\left[\begin{array}{rrr} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{array}\right]$$ Show that \(C^{2}=O\)

Find the inverse of each of the given matrices by using a calculator. $$\left[\begin{array}{rr} 20 & -45 \\ -12 & 24 \end{array}\right]$$

Solve the given systems of equations by determinants. Evaluate by expansion by minors. $$\begin{aligned} &2 x+y+z=4\\\ &2 y-2 z-t=3\\\ &3 y-3 z+2 t=1\\\ &6 x-y+t=0 \end{aligned}$$

Determine whether or not \(B=A^{-1}\) $$A=\left[\begin{array}{rrr} 1 & -1 & 3 \\ 3 & -4 & 8 \\ -2 & 3 & -4 \end{array}\right] \quad B=\left[\begin{array}{rrr} 8 & -5 & -4 \\ 4 & -2 & -1 \\ -1 & 1 & 1 \end{array}\right]$$
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