Question

Demonstrate the equation $\mathbf{\left|}\mathbf{d}\mathbf{e}\mathbf{t}\mathbf{A}\mathbf{\right|}\mathbf{=}\mathbf{||}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{||}\mathbf{||}\overrightarrow{\mathbf{v}}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{||}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{||}{\overrightarrow{\mathbf{v}}}_{\mathbf{n}}^{\mathbf{\perp }}\mathbf{||}$for a noninvertible$\mathit{n}\mathbf{\times }\mathit{n}$matrix$\mathit{A}\mathbf{=}\left[{\overrightarrow{v}}_1{\overrightarrow{v}}_1.........{\overrightarrow{v}}_n\right]$(Theorem 6.3.3).

Short answer

Since the columns of A are linearly dependent, this means that $\exists i\in \left\{1,.....,n\right\},{\overrightarrow{v}}_i^{\perp }=0$.

So the right side of this equation is also 0.

Step-by-step solution

Matrix Definition. 

Matrix is aset of numbers arrangedin rows and columns soas to form a rectangulararray.

The numbers are called the elements, or entries, of the matrix.

If there are m rows and n columns, the matrix is said to be a “ by ” matrix, written “m x n .”

To demonstrate the equation. 

If

$A=\left[{\overrightarrow{v}}_1{\overrightarrow{v}}_1......{\overrightarrow{v}}_n\right]\in {\mathrm{ℝ}}^{n\times n}$

Is a non-invertible matrix, then A = 0 .

Since the columns of are linearly dependent, this means that$\exists i\in \left\{1,....,n\right\}{\overrightarrow{v}}_i^{\perp }=0$.

So the right side of this equation is also 0.