Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Demonstrate Theorem 6.3.6 for linearly dependent vector ${\overset{⇀}{v}}_{1}, . . . ., {\overset{⇀}{v}}_{m}$ .

Therefore, the area of the parallelepiped is given by,

$\sqrt{d e t (A^{T} A)} = 0$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition.

Generally, the parallelepiped is a three-dimensional geometric solid with six faces that are parallelograms.

Parallelepiped is a three-dimensional solid shape.

It has 6 faces, 12 edges, and 8 vertices.

All faces of a parallelepiped are in the shape of a parallelogram.

Step 2: To demonstrate the linear dependent vectors.

If ${\overset{⇀}{v}}_{1}, {\overset{⇀}{v}}_{2}, . . . . ., {\overset{⇀}{v}}_{m}, \in ℝ^{2}$ are linearly dependent, then the matrix A that they form has linearly dependent columns.

Therefore, $A^{T} A$ is non-invertible.

So, the area of the parallelepiped is given by

$\sqrt{d e t (A^{T} A)} = 0$ .

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Linear Algebra With Applications

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Short Answer

Demonstrate Theorem 6.3.6 for linearly dependent vector ${\overset{⇀}{v}}_{1}, . . . ., {\overset{⇀}{v}}_{m}$ .

Step by Step Solution

Step 1: Definition.

Step 2: To demonstrate the linear dependent vectors.

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Linear Algebra With Applications

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Short Answer

Demonstrate Theorem 6.3.6 for linearly dependent vector v⇀1,....,v⇀m.

Step by Step Solution

Step 1: Definition.

Step 2: To demonstrate the linear dependent vectors.

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Demonstrate Theorem 6.3.6 for linearly dependent vector ${\overset{⇀}{v}}_{1}, . . . ., {\overset{⇀}{v}}_{m}$ .