This problem aims to find the volume of a **parallelepiped**, whose one vertex is at the origin **(0,0)** and the other **3** vertices are given. To solve this problem, it is required to have knowledge of **3-dimensional shapes** along with their** areas** and **volumes** and to calculate determinants of the **3×3** square matrix.

## Expert Answer

A **parallelepiped** is a 3-dimensional shape formed by six individual parallelograms. It is related to a **parallelogram** the same as a cube is related to a **square**.

To keep things simple, we will construct a** 3×3** matrix **A**, where the column entries are coordinates of the adjacent vertices of the given parallelepiped.

\[A=\left[\begin {matrix}1&-2&-1\\3& &3\\0&2&-1\\\end {matrix}\right]\]

The formula to find the volume is a dot product of the base of the parallelogram and its tilted altitude. But in matrix notation, the parallelepiped volume is equal to the absolute value of the determinant of $A$.

Volume = $|det(A)|$

Adjusting the matrix $A$ in the formula gives us:

\[volume=\left|\begin{matrix}1&-2&-1\\3&0&3\\0&2&-1\\\end{matrix}\right|\]

Next, we will solve for $det(A)$. Note that the determinant can only be found in a square matrix such as $A$.

We will find the determinant by using **co-factor expansion** across the first column.

\[=\left|\begin{matrix}0&3\\2&-1\\\end{matrix}\right|-3\left|\begin{matrix}-2& -1\\2& -1\\ \end {matrix} \right| +0 \left |\begin {matrix} -2 & -1\\ 0 & 3\\ \end {matrix} \right| \]

## Numerical Answer

Expanding the first column gives us only 2 entries as $a_13$ is equal to 0, but a complete solution is given here for simplicity.

\[ = [ (0)(-1) – (2)(3) ] + (-3)[ (-2)(-1) – (2)(-1) ] \]

\[ = -6 + (-3)[ 2 +2] \]

\[ = -6 + (-3)(4)\]

\[ = -6 + (-3)(4)\]

\[ = -6 – 12\]

\[ volume = -18 \]

Therefore, the volume of the given parallelepiped is equal to $18$.

### Example

**Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at $ (1, 0, -3), (1, 2, 4), (5, 1, 0)$.**

As the first step, we will construct a $3\times3$ matrix $A$, whose column entries are coordinates of the adjacent vertices of the given parallelepiped.

\[A = \left [\begin {matrix} 1 & 1 & 5 \\ 0 & 2 & 1\\ -3 & 4 & 0\\ \end {matrix} \right] \]

The volume of the parallelepiped can be calculated by taking absolute value of determinant of $A$.

\[ Volume = |det(A)| \]

Adjusting the matrix $A$ in the formula gives us:

\[ volume = \left |\begin {matrix} 1 & 1 & 5 \\ 0 & 2 & 1\\ -3 & 4 & 0\\ \end {matrix} \right| \]

Next, we will solve for $det(A)$ by using **co-factor expansion** across the first column.

\[ = \left |\begin {matrix} 2 & 1\\ 4 & 0\\ \end {matrix} \right| -(0) \left |\begin {matrix} 1 & 5\\ 4 & 0\\ \end {matrix} \right| +(-3) \left |\begin {matrix} 1 & 5\\ 2 & 1\\ \end {matrix} \right| \]

Equation becomes:

\[ v = -4+27 \]

\[ volume = 23 \]

Thus, the volume of parallelepiped comes out to be $23$.