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Expert-verified Found in: Page 306 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Demonstrate Theorem 6.3.6 for linearly dependent vector ${\stackrel{\mathbf{⇀}}{\mathbf{v}}}_{{\mathbf{1}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{\stackrel{\mathbf{⇀}}{\mathbf{v}}}_{{\mathbf{m}}}$.

Therefore, the area of the parallelepiped is given by,

$\sqrt{det\left({A}^{T}A\right)}=0$.

See the step by step solution

## 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 ${\stackrel{⇀}{v}}_{1},{\stackrel{⇀}{v}}_{2},.....,{\stackrel{⇀}{v}}_{m},\in {\mathrm{ℝ}}^{2}$ 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

$\sqrt{det\left({A}^{T}A\right)}=0$. ### Want to see more solutions like these? 