Question

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

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

It is given in the question that the determinant of the matrix is zero. Therefore,

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.