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 $23$.