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Linear Algebra and its Applications
Found in: Page 165
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Let \(u = \left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right]\), and \(v = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Compute the area of the parallelogram

determined by u, v, \({\bf{u}} + {\bf{v}}\), and 0, and compute the determinant of \(\left[ {\begin{array}{*{20}{c}}{\bf{u}}&{\bf{v}}\end{array}} \right]\). How do they compare? Replace the first entry of v by an arbitrary number x, and repeat the problem. Draw a picture and explain what you find.

The area of the parallelogram is 6 square units. The determinant of \(\left[ {\begin{array}{*{20}{c}}{\bf{u}}&{\bf{v}}\end{array}} \right]\) is 6. By replacing the first entry of v by an arbitrary number x, the area remains the same.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the vector                    

Obtain the vector by using the vectors and .

Thus, the vector is .

Step 2: Display the vectors on the graph

The graph of vectors , , , and 0 using arrows is shown below:

Here, the length of the base of the parallelogram is 3 units, and the height is 2 units.

The area of the parallelogram is calculated below:

Thus, the area is 6 square units.

Step 3: Compute the determinant

Obtain the determinant of the vector .

Thus, .

The area computed by the graph and the determinant of vectors is the same.

It means the sides of the parallelogram adjacent to 0 defines the side of the parallelogram, and it is equal to the area of the parallelogram.

Step 4: Replace the entry with an arbitrary number

Replace 1 by x in the vector to make it .

Obtain the vector using the vectors and .

Replace 1 by x in the vector to make it .

Obtain the vector using the vectors and .

Here, the length of the base of the parallelogram is 3 units, and the height is 2 units.

The area of the parallelogram is calculated below:

Thus, the area is 6 square units. The obtained area is the same as in the first case because the base and height of the parallelogram are not changing.

So, it does not affect the area.

Obtain the determinant of vectors .

Thus, .

Most popular questions for Math Textbooks

Question: In Exercise 10, determine the values of the parameter s for which the system has a unique solution, and describe the solution.

10.

\(\begin{array}{c}s{x_{\bf{1}}} - {\bf{2}}{x_{\bf{2}}} = {\bf{1}}\\4s{x_{\bf{1}}} + {\bf{4}}s{x_{\bf{2}}} = {\bf{2}}\end{array}\)

Question: 6. Use Cramer’s rule to compute the solution of the following system.

\(\begin{array}{c}{x_{\bf{1}}} + {\bf{3}}{x_{\bf{2}}} + \,{x_{\bf{3}}} = {\bf{4}}\\ - {x_{\bf{1}}} + \,\,\,\,\,\,\,\,\,\,{\bf{2}}{x_{\bf{3}}} = {\bf{2}}\\{\bf{3}}{x_{\bf{1}}} + \,{x_{\bf{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\, = {\bf{2}}\end{array}\)

Question: Use Cramer’s rule to compute the solutions of the systems in Exercises1-6.

4. \(\begin{array}{c} - 5{x_1} + 2{x_2} = 9\\3{x_1} - {x_2} = - 4\end{array}\)

Each equation in Exercises 1-4 illustrates a property of determinants. State the property

\(\left| {\begin{array}{*{20}{c}}{\bf{0}}&{\bf{5}}&{ - {\bf{2}}}\\{\bf{1}}&{ - {\bf{3}}}&{\bf{6}}\\{\bf{4}}&{ - {\bf{1}}}&{\bf{8}}\end{array}} \right| = - \left| {\begin{array}{*{20}{c}}{\bf{1}}&{ - {\bf{3}}}&{\bf{6}}\\{\bf{0}}&{\bf{5}}&{ - {\bf{2}}}\\{\bf{4}}&{ - {\bf{1}}}&{\bf{8}}\end{array}} \right|\)

Is it true that \(det \left( {A + B} \right) = det A + det B\)? Experiment with four pairs of random matrices as in Exercise 44, and make a conjecture.
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