Question

If a, b, c are unequal what is the condition that the value of the following determinant is zero.

\[\delta = \left( {\begin{array}{*{20}{c}}
  a&{{a^2}}&{{a^3} + 1} \\
  b&{{b^2}}&{{b^2} + 1} \\
  c&{2a}&{{c^3} + 1}
\end{array}} \right)\]
A.$1 + abc = 0$
B.$a + b + c + 1 = 0$
C.$(a - b)(b - c)(c - a) = 0$
D.None of these

Answer 1

Hint: It is given in the question that the determinant of the matrix is zero. Therefore,
\[ \Rightarrow \vartriangle = |\delta | = \left| {\begin{array}{*{20}{c}}
  a&{{a^2}}&{{a^3} + 1} \\
  b&{{b^2}}&{{b^3} + 1} \\
  c&{{c^2}}&{{c^3} + 1}
\end{array}} \right| = 0\]
Find the determinant of the matrix given in the question and equate it with zero and apply all the given options in the question.
Determinant of a matrix can be calculated by expanding the matrix either row wise or column wise. It is a single number and determinants are always calculated from a singular matrix.

Complete step-by-step answer:
We are given with a matrix whose determinant is equal to zero i.e.
 \[ \Rightarrow \vartriangle = |\delta | = \left| {\begin{array}{*{20}{c}}
  a&{{a^2}}&{{a^3} + 1} \\
  b&{{b^2}}&{{b^3} + 1} \\
  c&{{c^2}}&{{c^3} + 1}
\end{array}} \right| = 0\]
 To find the determinant expand it row wise, you can also expand it column wise as follows:
 \[ \Rightarrow \vartriangle = \left| {\begin{array}{*{20}{c}}
  a&{{a^2}}&{{a^3} + 1} \\
  b&{{b^2}}&{{b^3} + 1} \\
  c&{{c^2}}&{{c^3} + 1}
\end{array}} \right|\]

We can split this matrix as elements in the third column of the matrix consisting of two terms. So, we can express them in addition of two matrices as follows:

$ \Rightarrow \vartriangle = \left| {\begin{array}{*{20}{c}}
  a&{{a^2}}&{{a^3}} \\
  b&{{b^2}}&{{b^3}} \\
  c&{{c^2}}&{{c^3}}
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  a&{{a^2}}&1 \\
  b&{{b^2}}&1 \\
  c&{{c^2}}&1
\end{array}} \right|$
Taking a, b & c as common from row 1, row 2 & row 3 respectively, we get
 $ \Rightarrow \vartriangle = abc\left| {\begin{array}{*{20}{c}}
  1&a&{{a^2}} \\
  1&b&{{b^2}} \\
  1&c&{{c^2}}
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  a&{{a^2}}&1 \\
  b&{{b^2}}&1 \\
  c&{{c^2}}&1
\end{array}} \right|$
 Taking the matrix as common as both the matrices are same, we get
 $ \Rightarrow \vartriangle = (1 + abc)\left| {\begin{array}{*{20}{c}}
  a&{{a^2}}&1 \\
  b&{{b^2}}&1 \\
  c&{{c^2}}&1
\end{array}} \right|$
 It is given that the determinant of this matrix is zero. Therefore,
 $ \Rightarrow (abc + 1)\left| {\begin{array}{*{20}{c}}
  a&{{a^2}}&1 \\
  b&{{b^2}}&1 \\
  c&{{c^2}}&1
\end{array}} \right| = 0$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  a&{{a^2}}&1 \\
  b&{{b^2}}&1 \\
  c&{{c^2}}&1
\end{array}} \right| = 0$ or $1 + abc = 0$
 As the matrix cannot be equal to zero because a, b and c are not equal. Hence, $1 + abc = 0$ is the required condition.
 So, A is the correct option

Note: In the given question, students get confused in applying the property of determinants. According to that property, if each element in a row or column is expressed as the sum of two terms or more terms then the determinants can also be expressed as a sum of two or more determinants.