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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Prove that:
$\begin{vmatrix}\text{a}^2&\text{bc}&\text{ac}+\text{c}^2\\\text{a}^2+\text{ab}&\text{b}^2&\text{ac}\\\text{ab}&\text{b}^2+\text{ac}&\text{c}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{a}^2&\text{bc}&\text{ac}+\text{c}^2\\\text{a}^2+\text{ab}&\text{b}^2&\text{ac}\\\text{ab}&\text{b}^2+\text{ac}&\text{c}^2\end{vmatrix}$
$=\text{abc}\begin{vmatrix}\text{a}&\text{c}&\text{a}+\text{c}\\\text{a}+\text{b}&\text{b}&\text{a}\\\text{b}&\text{b}+\text{c}&\text{c}\end{vmatrix}$
[Taking out a, b and c common from $C_1, C_2$ and $C_3$​​​​​​​]
$=\text{abc}\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}+\text{b}&\text{b}&-2\text{b}\\\text{b}&\text{b}+\text{c}&-2\text{b}\end{vmatrix}$
[Applying $C_3→ C_3 - C_2 - C_1$]
$=(\text{abc})(-2\text{b})\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}+\text{b}&\text{b}&1\\\text{b}&\text{b}+\text{c}&1\end{vmatrix}$
[Taking (-2b) common from $C_3​​​​​​​$]
$=(\text{abc})(-2\text{b})\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}&-\text{c}&0\\\text{b}&\text{b}+\text{c}&1\end{vmatrix}$
[Applying $R_2 → R_2 - R_1​​​​​​​$​​​​​​​]
$=(\text{abc})(-2\text{b})\times1\begin{vmatrix}\text{a}&\text{c}\\\text{a}&-\text{c}\end{vmatrix}$
[expanding along $C_3$​​​​​​​]
$=(\text{abc})(-2\text{b})(-2\text{ac})$
$=4\text{a}^2\text{b}^2\text{c}^2$
$=\text{R.H.S}$

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