Question

The determinant of $\left| \begin{array}{l} 1 \;\;\;\;\;\;\;\; 1 \;\;\;\;\;\;\;\; 1 \\\\{\alpha ^2}{\text{ }}\;\;\;\;\;{\beta ^2}{\text{ }}\;\;\;\;\;{\gamma ^2}\\\ - {\alpha ^2}{\text{ }}\;\;\;{\beta ^2}{\text{ }}\;\;\;{\gamma ^2}\end{array} \right|$ is divisible by?
A) $\beta - \gamma$
B) $\alpha + \beta$
C) $\beta \gamma$
D) $\dfrac{\beta }{\gamma }$

Answer 1

Here we will find the value of the determinant using the formula for the value of determinant. Then we will factorize the value of the determinant that we have obtained. We will then verify if any of the options given in the question is a factor of the determinant. If we find any match, then the determinant will be divisible by the term given in the question.

Complete step by step solution:
We will first calculate the value of the given determinant.
$D = \left| \begin{array}{l} 1 \;\;\;\;\;\;\;\; 1 \;\;\;\;\;\;\;\;\;1 \\\\{\alpha ^2}{\text{ }}\;\;\;\;\;\;{\beta ^2}{\text{ }}\;\;\;\;\;\;{\gamma ^2}\\\ - {\alpha ^2}{\text{ }}\;\;\;\;{\beta ^2}{\text{ }}\;\;\;\;{\gamma ^2}\end{array} \right|$
$\Rightarrow D = 1\left( {{\beta ^2}{\gamma ^2} - {\gamma ^2}{\beta ^2}} \right) - 1\left( {{\alpha ^2}{\gamma ^2} - {\gamma ^2}\left( { - {\alpha ^2}} \right)} \right) + 1\left( {{\alpha ^2}{\beta ^2} - {\beta ^2}\left( { - {\alpha ^2}} \right)} \right)$
Multiplying the terms, we get
$\Rightarrow D = {\beta ^2}{\gamma ^2} - {\gamma ^2}{\beta ^2} - {\alpha ^2}{\gamma ^2} - {\gamma ^2}{\alpha ^2} + {\alpha ^2}{\beta ^2} + {\beta ^2}{\alpha ^2}$
On adding and subtracting the like terms, we get
$\Rightarrow D = - 2{\gamma ^2}{\alpha ^2} + 2{\alpha ^2}{\beta ^2}$
On further simplification, we get
$\Rightarrow D = 2{\alpha ^2}\left( {{\beta ^2} - {\gamma ^2}} \right)$
Using the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, we get
$\Rightarrow D = 2{\alpha ^2}\left( {\beta + \gamma } \right)\left( {\beta - \gamma } \right)$
Thus, we can say that $\beta - \gamma$ is one of the factors of the determinant. Thus, the determinant of $\left| \begin{array}{l} 1 \;\;\;\;\;\; 1 \;\;\;\;\;\; 1 \\\\{\alpha ^2}{\text{ }}\:\;\;{\beta ^2}{\text{ }}\;\;\;{\gamma ^2} \\\ - {\alpha ^2}{\text{ }}\;{\beta ^2}{\text{ }}\;{\gamma ^2}\end{array} \right|$ is divisible by $\beta - \gamma$.

Hence, the correct option is option A.

Note:
Factor of any number is defined as the number which when multiplied by another number gives us the original number. The most important property of a determinant is that if two or more rows are similar then the value of the determinant is equal to zero. Also, if we interchange the rows and columns the value of determinant remains the same. If all the elements of any one of the rows or columns are zero then the value of the determinant is equal to zero. If we multiply all the elements of a row by a constant then the resultant determinant will be constant times the value of the original determinant.