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Determinants — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Evaluate:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix}$
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Answer

Let $\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix}$
Applying $C_1 \rightarrow C_1 + C_2 + C_3$ we get,
$\triangle=\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{b}&\text{c}\\\text{a}+\text{b}+\text{c}&\text{a}&\text{b}\\\text{a}+\text{b}+\text{c}&\text{c}&\text{a}\end{vmatrix}$
Taking (a + b + c) common, we have
$\triangle=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\1&\text{a}&\text{b}\\1&\text{c}&\text{a}\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1, R_3 - R_1​​​​​​​$​​​​​​​, we get
$\triangle=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\0&\text{a}-\text{b}&\text{b}-\text{c}\\0&\text{c}-\text{b}&\text{a}-\text{c}\end{vmatrix}$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})[(\text{a}-\text{b})(\text{a}-\text{c})-(\text{b}-\text{c})(\text{c}-\text{b})]$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})\big[\text{a}^2-\text{ac}-\text{ab}+\text{bc}+\text{b}^2+\text{c}^2-2\text{ab}\big]$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})\big[\text{a}^2+\text{b}^2+\text{c}^2-\text{ac}-\text{ab}-\text{bc}\big]$

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