Question

If $ \left| {\begin{matrix}
   a & {{a^2}} & {1 + {a^3}} \\
   b & {{b^2}} & {1 + {b^3}} \\
   c & {{c^2}} & {1 + {c^3}} \\
 \end{matrix} } \right| = 0 $ and the vectors A =$\left( {1,a,{a^2}} \right)$, B =$\left( {1,b,{b^2}} \right)$, C =$\left( {1,c,{c^2}} \right)$are non-coplanar, then find the value of the product rule abc.
A. 0
B. 1
C. -1
D. None of these

Answer 1

Hint: Using the definition of non-coplanar vectors, it is known that$\Delta = \left| {\begin{matrix}
   1 & a & {{a^2}} \\
   1 & b & {{b^2}} \\
   1 & c & {{c^2}} \\

 \end{matrix} } \right| \ne 0$. That is their scalar triple product is not equal to zero. And using matrix regular splitting, we will split the given matrix $ \left| {\begin{matrix}
   a & {{a^2}} & {1 + {a^3}} \\
   b & {{b^2}} & {1 + {b^3}} \\
   c & {{c^2}} & {1 + {c^3}} \\

 \end{matrix} } \right| $ into two matrices. This provides us to find the value the ‘abc’ by taking the common matrix from the two matrices.

Complete step-by-step answer:
Since$\left( {1,a,{a^2}} \right)$,$\left( {1,b,{b^2}} \right)$,$\left( {1,c,{c^2}} \right)$are non-coplanar
$\Delta = \left| {\begin{matrix}
   1 & a & {{a^2}} \\
   1 & b & {{b^2}} \\
   1 & c & {{c^2}} \\

 \end{matrix} } \right| \ne 0$… (1)
Since the three products are non-coplanar, their scalar triple product is not equal to zero. That is simply their determinant is not zero.
If two vectors (their carrier lines) don’t intersect, there is no common plane, so the vectors (and also the lines) are said to be non-coplanar.
Given
 $ \left| {\begin{matrix}
   a & {{a^2}} & {1 + {a^3}} \\
   b & {{b^2}} & {1 + {b^3}} \\
   c & {{c^2}} & {1 + {c^3}} \\

 \end{matrix} } \right| = 0 $
The determinant is a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the linear transformation described by the matrix. The determinant of a matrix A is denoted det (A), det A or$\left| {\text{A}} \right|$.
 $ \Rightarrow \left| {\begin{matrix}
   a & {{a^2}} & 1 \\
   b & {{b^2}} & 1 \\
   c & {{c^2}} & 1 \\

 \end{matrix} } \right| + \left| {\begin{matrix}
   a & {{a^2}} & {{a^3}} \\
   b & {{b^2}} & {{b^3}} \\
   c & {{c^2}} & {{c^3}} \\

 \end{matrix} } \right| = 0 $
A matrix splitting is an expression which represents a given matrix as a sum or difference of matrices. We have regular splitting’s and matrix iterative methods for splitting of matrices.
 $ \Rightarrow \left( {1 + abc} \right)\left| {\begin{matrix}
   1 & a & {{a^2}} \\
   1 & b & {{b^2}} \\
   1 & c & {{c^2}} \\

 \end{matrix} } \right| = 0 $
We had now taken common value from the both matrices.
 $ \Rightarrow \left( {1 + abc} \right)\Delta = 0 $
From (1) $ \Delta \ne 0 $ , we get
1 + $abc$= 0
$ \Rightarrow abc = - 1$

So, the correct answer is “Option C”.

Note: We can solve the problems on coplanar vectors in the same way, where the three vectors are coplanar if their scalar triple product is zero. The three vectors are coplanar if they are linearly dependent. For n vectors, vectors are coplanar if among them no more than two linearly independent vectors.