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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Prove that $\begin{vmatrix}\text{bc}-\text{a}^2&\text{ca}-\text{b}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2&\text{ab}-\text{c}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2&\text{bc}-\text{a}^2&\text{ca}-\text{b}^2\end{vmatrix}$ is divisible by (a + b + c) and find the quotient.
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Answer

$\Delta=\begin{vmatrix}\text{bc}-\text{a}^2&\text{ca}-\text{b}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2&\text{ab}-\text{c}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2&\text{bc}-\text{a}^2&\text{ca}-\text{b}^2\end{vmatrix}$
$\big[\text{Applying C}_1\rightarrow\text{C}_1-\text{C}_2\text{ and C}_2\rightarrow\text{C}_2-\text{C}_3\big]$
$\Delta=\begin{vmatrix}\text{bc}-\text{a}^2-\text{ca}+\text{b}^2&\text{ca}-\text{b}^2-\text{ab} +\text{c}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2-\text{ab}+\text{c}^2&\text{ab}-\text{c}^2-\text{bc}+\text{a}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2-\text{bc}+\text{a}^2&\text{bc}-\text{a}^2-\text{ca}+\text{b}^2&\text{ca}-\text{b}^2\\\end{vmatrix}$
$=\begin{vmatrix}(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})&(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})&\text{ab}-\text{c}^2\$\text{c}-\text{b})(\text{a}+\text{b}+\text{c})&(\text{a}-\text{c})(\text{a}+\text{b}+\text{c})&\text{bc}-\text{a}^2\$\text{a}-\text{c})(\text{a}+\text{b}+\text{c})&(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})&\text{ca}-\text{b}^2\end{vmatrix}$
$\big[\text{Taking }(\text{a}+\text{b}+\text{c})\text{ common from C}_1\text{ and C}_2\text{ each}\big]$
$\Delta=(\text{a}+\text{b}+\text{c})=\begin{vmatrix}\text{b}-\text{a}&\text{c}-\text{b} &\text{ab}-\text{c}^2\\\text{c}-\text{b}&\text{a}-\text{c}&\text{bc}-\text{a}^2\\\text{a}-\text{c}&\text{b}-\text{a}&\text{ca}-\text{b}^2\end{vmatrix}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3\big]$
$\Delta=(\text{a}+\text{b}+\text{c})=\begin{vmatrix}0&0&\text{ab}+\text{bc}+\text{ca}-(\text{a}^2+\text{b}^2+\text{c}^2)\\\text{c}-\text{b}&\text{a}-\text{c}&\text{bc}-\text{a}^2\\\text{a}-\text{c}&\text{b}-\text{a}&\text{ca}-\text{b}^2\end{vmatrix}$
[Expanding along $R_1$]
$\Delta=(\text{a}+\text{b}+\text{c})^2\big[\text{ab}+\text{bc}+\text{ca}-(\text{a}^2+\text{b}^2+\text{c}^2)\big]\big[(\text{c}-\text{b})(\text{b}-\text{a})(\text{a}-\text{c})^2\big]$
$=(\text{a}+\text{b}+\text{c})^2(\text{ab}+\text{bc}+\text{ca}-\text{a}^2-\text{b}^2-\text{c}^2)\times(\text{bc}-\text{ac}-\text{b}^2+\text{ab}-\text{a}^2-\text{c}^2+2\text{ac})$
$=(\text{a}+\text{b}+\text{c})\big[(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})^2\big]$
Hence, given deteminant is divisible by (a + b + c) and quotient is
$(\text{a}+\text{b}+\text{c})\big(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\big)^2$

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