Question

Find the area (or 2-volume) of the parallelogram (or 2parallelepiped) defined by the vectors

$\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]\mathit{A}\mathit{n}\mathit{d}\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$

Short answer

Therefore, the area of the parallelogram of the vectors is given by,

$\sqrt{det\left(A^{T}A\right)}=\sqrt{20}$

Step-by-step solution

Definition. 

The area of a parallelogram is the number of square units inside the polygon.

The area of a parallelogram can be determined by multiplying the base and height.

For finding the area of a parallelogram, the base and height must be perpendicular. The area of a parallelogram is given by,

Area = b x h

Given. 

Given Matrix,

$\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]$And $\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$

To find area of the parallelogram of the vectors.

These two vectors form the matrix

$A=\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\\ 1& 4\end{array}\right]$.

Therefore, the area is

$$\begin{gathered} \sqrt{det\left(A^{T}A\right)}=\sqrt{det\left(\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 3& 4\end{array}\right]\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\\ 1& 4\end{array}\right]\right)} \\ \sqrt{det\left(A^{T}A\right)}=\sqrt{det\left(\left[\begin{array}{cc}4& 10\\ 10& 30\end{array}\right]\right)} \\ \sqrt{det\left(A^{T}A\right)}=\sqrt{20} \end{gathered}$$