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Chapter 6: Problem 66

What is the fastest method for solving a linear system with your graphing utility?

Short Answer

Expert verified
The fastest method for solving a linear system with a graphing utility involves identifying and plotting the system of equations and finding their points of intersection, which represent the solutions of the system.

Step by step solution

01

Identify the System of Equations

Identify the system of linear equations that need to be solved. This system of equations will be provided in the problem statement.
02

Plot the Equations

Use your graphing utility to plot the equations. Each equation will be a separate line on the graph.
03

Find the Points of Intersection

Using the graphing utility, find the points where the lines intersect. These points are the solutions to the system of equations. If the lines intersect at one point, then there is one solution for that linear system. If the lines are parallel (they do not intersect), then there are no solutions. If the lines coincide (are on top of each other), then there are infinite solutions.

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Most popular questions from this chapter

Explain how to find the multiplicative inverse for a \(3 \times 3\) invertible matrix.

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$\text { If } A=\left[\begin{array}{ll}3 & 5 \\\2 & 4\end{array}\right], \text { find}\left(A^{-1}\right)^{-1}$$

Use a graphing utility to find the multiplicative inverse of each matrix. Check that the displayed inverse is correct. $$\left[\begin{array}{rr}3 & -1 \\\\-2 & 1\end{array}\right]$$

In Exercises \(37-44\), perform the indicated matrix operations given that \(A, B,\) and \(C\) are defined as follows If an operation is not defined, state the reason. $$A=\left[\begin{array}{rr}4 & 0 \\\\-3 & 5 \\\0 & 1\end{array}\right] \quad B=\left[\begin{array}{rr} 5 & 1 \\\\-2 & -2\end{array}\right] \quad C=\left[\begin{array}{rr}1 & -1 \\\\-1 & 1\end{array}\right]$$ $$A(B+C)$$

Determinants are used to find the area of a triangle whose vertices are given by three points in a rectangular coordinate system. The area of a triangle with vertices \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\) and \(\left(x_{3}, y_{3}\right)\) is $$\text { Area }-\pm \frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\\x_{2} & y_{2} & 1 \\\x_{3} & y_{3} & 1\end{array}\right|$$ where the \(\pm\) symbol indicates that the appropriate sign should be chosen to yield a positive area. Use determinants to find the area of the triangle whose vertices are \((1,1),(-2,-3),\) and \((11,-3)\)
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