Question

The determinant of \left| {\begin{array}{*{20}{c}}1&1&1\\\\{{\alpha ^2}}&{{\beta ^2}}&{{\gamma ^2}}\\\\{ - {\alpha ^2}}&{{\beta ^2}}&{{\gamma ^2}}\end{array}} \right| is divisible by?
A.$\beta - \gamma$
B.$\alpha + \beta$
C.$\beta \gamma$
D.$\beta /\gamma$

Answer 1

We will find the value of the determinant using the formula for the value of the determinant. We will factorize the value of the determinant that we have found. Then we will 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 option.

Formulas used:
We will use the following formulas:
The formula for the value of the determinant D = \left| {\begin{array}{*{20}{c}}a&b;&c;\\\d&e;&f;\\\g&h;&i;\end{array}} \right| is given by$D = a\left( {ei - fh} \right) - b\left( {di - fg} \right) + c\left( {dh - eg} \right)$.
The difference of 2 square numbers is given by: ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$.

Complete step-by-step answer:
We will first calculate the determinant. We will substitute 1 for $a$, 1 for $b$, 1 for $c$, ${\alpha ^2}$ for $d$, ${\beta ^2}$ for $e$, ${\gamma ^2}$ for $f$, $- {\alpha ^2}$ for $g$, ${\beta ^2}$ for $h$ and ${\gamma ^2}$ for $i$ in the formula for the value of the determinant $D$.
\begin{array}{l}D = \left| {\begin{array}{*{20}{c}}1&1&1\\\\{{\alpha ^2}}&{{\beta ^2}}&{{\gamma ^2}}\\\\{ - {\alpha ^2}}&{{\beta ^2}}&{{\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)\end{array}
We will simplify the above equation. First, we will simplify the expressions inside the brackets:
$\begin{array}{l} \Rightarrow D = 1\left( {{\beta ^2}{\gamma ^2} - {\gamma ^2}{\beta ^2}} \right) - 1\left( {{\alpha ^2}{\gamma ^2} - {\gamma ^2}\left( { - {\alpha ^2}} \right)} \right) + 1\left( {{\alpha ^2}{\beta ^2} - {\beta ^2}\left( { - {\alpha ^2}} \right)} \right)\\\ \Rightarrow D = 1\left( 0 \right) - 1\left( {{\alpha ^2}{\gamma ^2} + {\alpha ^2}{\gamma ^2}} \right) + 1\left( {{\alpha ^2}{\beta ^2} + {\alpha ^2}{\beta ^2}} \right)\end{array}$
Adding the terms inside the brackets, we get
$\Rightarrow D = 1\left( 0 \right) - 1\left( {2{\alpha ^2}{\gamma ^2}} \right) + 1\left( {2{\alpha ^2}{\beta ^2}} \right)$
Solving the brackets, we get
$\begin{array}{l} \Rightarrow D = 0 - 2{\alpha ^2}{\gamma ^2} + 2{\alpha ^2}{\beta ^2}\\\ \Rightarrow D = 2{\alpha ^2}{\beta ^2} - 2{\alpha ^2}{\gamma ^2}\end{array}$
We need to find out whether the determinant is divisible by any of the terms given in the options. For this, we will factorize the determinant. We will take out the term $2{\alpha ^2}$ as it is common to both the terms:
$\Rightarrow D = 2{\alpha ^2}\left( {{\beta ^2} - {\gamma ^2}} \right)$
We will substitute $\beta$ for $a$ and $\gamma$ for $b$ in the formula for the difference of 2 square numbers:
$\Rightarrow D = 2{\alpha ^2}\left( {\beta + \gamma } \right)\left( {\beta - \gamma } \right)$
We can see from the above equation that $\beta - \gamma$ is a divisor of the determinant.
$\therefore$ Option A is the correct option.

Note: We know that a factor of any number say $n$ is any number (or term) which when multiplied with another number (or term) gives us $n$. Here we have found out the determinant. It is important for us to remember that if two rows or two columns are equal then the determinant will be 0.