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For each of the following linear systems, obtain a solution by graphical methods, if possible. Explain the results from a geometrical standpoint. a. \(\quad x_{1}+2 x_{2}=0\) b. \(\quad x_{1}+2 x_{2}=3\) c. \(\quad 2 x_{1}+x_{2}=-1\) d. \(\quad 2 x_{1}+x_{2}+x_{3}=1\), \(x_{1}-x_{2}=0\) \(-2 x_{1}-4 x_{2}=6\) \(x_{1}-3 x_{2}=5\)

Short Answer

Expert verified
The solutions for the linear equations are: a. \(x_{1}= any\), \(x_{2}= -0.5*x_{1}\); b. \(x_{1}= 2\), \(x_{2}= -1\); c. No solution; d. Infinite solutions.

Step by step solution

01

Graph the First Line

The first equation can be rearranged into slope-intercept form as \(x_{2}= -0.5*x_{1}\). Plot this on a graph anywhere along the x-axis where \(x_{2}\) equals zero.
02

Graph the Second Line

The second equation can be rearranged into slope-intercept form as \(x_{2}= 1.5 - 0.5*x_{1}\). Plot this on a graph where \(x_{2}\) equals 1.5 when \(x_{1}\) equals zero. Superimpose this graph onto the first.
03

Interpret the Graphs

The first and second graph intersect at one point, which is the solution. In this case, the solution is \(x_{1}= 2\), \(x_{2}= -1\). Any other solution would mean the two lines would have to intersect at a different point, which is not likely in simple 2D or 3D linear equation systems.
04

Graph the Third and Fourth Lines

Proceed in a similar manner for the third and fourth line, i.e, change them into slope-intercept form as \(x_{2}= -2*x_{1} - 1\) and \(x_{2}= x_{1}\), \(x_{3}= 1-2*x_{1}-x_{2}\), and superimpose these graphs onto earlier ones to check if there's any intersection.
05

Conclude the Results

If the graphs do not intersect at any point, it suggests that there are no solutions to the system. If the graphs intersect at a single point, that point is the solution to the system. If the lines coincide, it means there are infinitely many solutions

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Key Concepts

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

Linear Equations
A linear equation is an equation that creates a straight line when it is graphed. It generally follows a form like \(ax + by = c\) in two variables, where \(a\), \(b\), and \(c\) are constants. Linear equations are simple yet powerful tools for solving many algebraic problems.
In this context, let's talk more about these systems:
  • The equation \(x_{1} + 2x_{2} = 0\) and \(x_{1} + 2x_{2} = 3\) have two variables - \(x_{1}\) and \(x_{2}\).
  • These equations represent lines when graphed on a plane.
  • The solution to these equations is typically where these lines intersect if you graph them together.
This approach forms the foundation for understanding how we can use graphical methods to solve a system of linear equations. As we go further, we'll see how these graphs lead us into more complex discussions involving geometry and graphical representation.
Solution Interpretation
When finding solutions through graphical methods, interpretation is key. After plotting the lines of the linear equations on a cartesian plane:
  • Each point of intersection represents a potential solution to the system.
  • If two lines intersect at a single point, that particular point is the unique solution of the system.
  • If lines are parallel and never intersect, the system has no solution.
  • If they overlap perfectly, there are infinitely many solutions, as every point on the line is a solution.
This gives a visual and intuitive way to understand solutions. In our exercise, when we plot and intersect \(x_{1} + 2x_{2} = 0\) and \(x_{1} + 2x_{2} = 3\), we find a distinctive solution on their intersection, which further grounds our understanding through geometric perspective.
Graphical Representation
Graphical representation of linear equations brings to life the concept of solving linear systems. When each equation is graphed as a line, the interaction of these lines on the graph tells us about potential solutions.
  • Graphical representation simplifies complex algebraic expressions by focusing on visuals.
  • It provides a means to quickly estimate solutions by identifying intersections on the graph.
  • It can help illustrate concepts of no solution, unique solutions, and infinite solutions visually.
In a system containing two or more equations like \(x_{1}+2x_{2}=0\) and \(x_{1}+2x_{2}=3\), it is particularly effective. By translating algebraic equations into lines, students find themselves better equipped to interpret solutions through these visualizations rather than formulas alone.
Geometrical Approach
The geometrical approach to solving linear systems enhances understanding by focusing on spatial relationships. By interpreting the equations as geometric lines on a plotted graph, several interesting points arise:
  • Each linear equation translates into a line on a two-dimensional plane.
  • The point where lines intersect is potentially the set \( (x_1, x_2) \) that satisfies all equations in the system.
  • By understanding the geometrical premise, it is easier to grasp reasons behind having no solution (parallel lines) or infinitely many (coinciding lines).
Grasping solutions through a geometrical approach allows us to visualize relationships and solve equations by utilizing concepts of distance and angles fundamentally. Emphasizing these spatial relationships fosters a deeper understanding of both the mechanics and logic behind solving linear systems graphically.

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Most popular questions from this chapter

Use Gaussian elimination and three-digit chopping arithmetic to solve the following linear systems, and compare the approximations to the actual solution. a. \(\quad \begin{aligned} 0.03 x_{1}+58.9 x_{2} &=59.2 \\ 5.31 x_{1}-6.10 x_{2} &=47.0 . \end{aligned}\) Actual solution \([10,1]\). b. $$ \begin{aligned} 3.03 x_{1}-12.1 x_{2}+14 x_{3} &=-119 \\ -3.03 x_{1}+12.1 x_{2}-7 x_{3} &=120 \\ 6.11 x_{1}-14.2 x_{2}+21 x_{3} &=-139 \end{aligned} $$ Actual solution \(\left[0,10, \frac{1}{7}\right]\). c. \(1.19 x_{1}+2.11 x_{2}-100 x_{3}+x_{4}=1.12\), $$ 14.2 x_{1}-0.122 x_{2}+12.2 x_{3}-x_{4}=3.44 $$ $$ \begin{aligned} 100 x_{2}-99.9 x_{3}+x_{4} &=2.15 \\ 15.3 x_{1}+0.110 x_{2}-13.1 x_{3}-x_{4} &=4.16 \end{aligned} $$ Actual solution \([0.176,0.0126,-0.0206,-1.18]\) d. $$ \begin{aligned} &\pi x_{1}-e x_{2}+\sqrt{2} x_{3}-\sqrt{3} x_{4}=\sqrt{11} \\ &\pi^{2} x_{1}+e x_{2}-e^{2} x_{3}+\frac{3}{7} x_{4}=0 \\ &\sqrt{5} x_{1}-\sqrt{6} x_{2}+x_{3}-\sqrt{2} x_{4}=\pi \\ &\pi^{3} x_{1}+e^{2} x_{2}-\sqrt{7} x_{3}+\frac{1}{9} x_{4}=\sqrt{2} \end{aligned} $$ Actual solution \([0.788,-3.12,0.167,4.55]\)

Show that the operations a. \(\quad\left(\lambda E_{i}\right) \rightarrow\left(E_{i}\right)\) b. \(\quad\left(E_{i}+\lambda E_{j}\right) \rightarrow\left(E_{i}\right)\) c. \(\quad\left(E_{i}\right) \leftrightarrow\left(E_{j}\right)\) do not change the solution set of a linear system.

Suppose \(m\) linear systems $$A \mathbf{x}^{(p)}=\mathbf{b}^{(p)}, \quad p=1,2, \ldots, m$$ are to be solved, each with the \(n \times n\) coefficient matrix \(A\). a. Show that Gaussian elimination with backward substitution applied to the aug- mented matrix $$\left[\begin{array}{ll} A: & \mathbf{b}^{(1)} \mathbf{b}^{(2)} \cdots \mathbf{b}^{(m)} \end{array}\right]$$ requires $$\frac{1}{3} n^{3}+m n^{2}-\frac{1}{3} n \quad \text { multiplications/ divisions }$$ $$\frac{1}{3} n^{3}+m n^{2}-\frac{1}{2} n^{2}-m n+\frac{1}{6} n \quad \text { additions/subtractions. }$$ b. Show that the Gauss-Jordan method (see Exercise 12 , Section \(6.1\) ) applied to the augmented matrix $$\left[\begin{array}{ll} A: & \left.{\mathbf{b}^{(1)} \mathbf{b}^{(2)}} \cdots \mathbf{b}^{(m)}\right] \end{array}\right.$$ requires $$\frac{1}{2} n^{3}+m n^{2}-\frac{1}{2} n \quad \text { multiplications/divisions }$$ and $$\frac{1}{2} n^{3}+(m-1) n^{2}+\left(\frac{1}{2}-m\right) n \quad \text { additions/subtractions. }$$ c. For the special case $$\mathbf{b}^{(p)}=\left[\begin{array}{c} 0 \\ \vdots \\ 0 \\ 1 \\ \vdots \\ 0 \end{array}\right] \leftarrow p \text { th row }$$ for each \(p=1, \ldots, m\), with \(m=n\), the solution \(x^{(p)}\) is the \(p\) th column of \(A^{-1}\). Show that Gaussian elimination with backward substitution requires \(\frac{4}{3} n^{3}-\frac{1}{3} n \quad\) multiplications/divisions and $$\frac{4}{3} n^{3}-\frac{3}{2} n^{2}+\frac{1}{6} n \quad \text { additions/subtractions }$$ for this application, and that the Gauss-Jordan method requires $$\frac{3}{2} n^{3}-\frac{1}{2} n \quad \text { multiplications/divisions }$$ and $$\frac{3}{2} n^{3}-2 n^{2}+\frac{1}{2} n \quad \text { additions/subtractions. }$$ d. Construct an algorithm using Gaussian elimination to find \(A^{-1}\), but do not per- form multiplications when one of the multipliers is known to be 1 , and do not per- form additions/subtractions when one of the elements involved is known to be \(0 .\) Show that the required computations are reduced to \(n^{3}\) multiplications/divisions and \(n^{3}-2 n^{2}+n\) additions/subtractions. e. Show that solving the linear system \(A x=\mathbf{b}\), when \(A^{-1}\) is known, still requires \(n^{2}\) multiplications/divisions and \(n^{2}-n\) additions/subtractions. f. Show that solving \(m\) linear systems \(A x^{(p)}=\mathbf{b}^{(p)}\), for \(p=1,2, \ldots, m\), by the method \(x^{(p)}=\) \(A^{-1} \mathbf{b}(p)\) requires \(m n^{2}\) multiplications and \(m\left(n^{2}-n\right)\) additions, if \(A^{-1}\) is known. g. Let \(\mathrm{A}\) be an \(n \times n\) matrix. Compare the number of operations required to solve \(n\) linear systems involving \(A\) by Gaussian elimination with backward substitution and by first inverting \(A\) and then multiplying \(A x=\mathbf{b}\) by \(A^{-1}\), for \(n=3,10,50,100 .\) Is it ever advantageous to compute \(A^{-1}\) for the purpose of solving linear systems?

Let \(A\) be the \(10 \times 10\) tridiagonal matrix given by \(a_{i i}=2, a_{i, i+1}=a_{i, i-1}=-1\), for each \(i=2, \cdots, 9\), and \(a_{11}=a_{10,10}=2, a_{12}=a_{10,9}=-1\). Let \(\mathbf{b}\) be the ten-dimensional column vector given by \(b_{1}=b_{10}=1\) and \(b_{i}=0\), for each \(i=2,3, \cdots, 9\). Solve \(A \mathbf{x}=\mathbf{b}\) using the Crout factorization for tridiagonal systems.

a. Show that the \(L U\) Factorization Algorithm requires $$ \frac{1}{3} n^{3}-\frac{1}{3} n \text { multiplications/divisions and } \frac{1}{3} n^{3}-\frac{1}{2} n^{2}+\frac{1}{6} n \text { additions/subtractions. } $$ b. Show that solving \(L \mathbf{y}=\mathbf{b}\), where \(L\) is a lower- triangular matrix with \(l_{i i}=1\) for all \(i\), requires \(\frac{1}{2} n^{2}-\frac{1}{2} n\) multiplications/divisions and \(\frac{1}{2} n^{2}-\frac{1}{2} n\) additions/subtractions. c. Show that solving \(A \mathbf{x}=\mathbf{b}\) by first factoring \(A\) into \(A=L U\) and then solving \(L \mathbf{y}=\mathbf{b}\) and \(U \mathbf{x}=\mathbf{y}\) requires the same number of operations as the Gaussian Elimination Algorithm 6.1. d. Count the number of operations required to solve \(m\) linear systems \(A \mathbf{x}^{(k)}=\mathbf{b}^{(k)}\) for \(k=1, \ldots, m\) by first factoring \(A\) and then using the method of part (c) \(m\) times.

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