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Determinants — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDeterminants5 Marks
Question
Evaluate: $\begin{vmatrix}\text{a}&\text{b}+\text{c}&\text{a}^2\\\text{b}&\text{c}+\text{a}&\text{b}^2\\\text{c}&\text{a}+\text{b}&\text{c}^2\end{vmatrix}$
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Answer

$\begin{vmatrix}\text{a}&\text{b}+\text{c}&\text{a}^2\\\text{b}&\text{c}+\text{a}&\text{b}^2\\\text{c}&\text{a}+\text{b}&\text{c}^2\end{vmatrix}$
Apply$: C_2 \rightarrow C_2 + C_1$
$=\begin{vmatrix}\text{a}&\text{b}+\text{c}+\text{a}&\text{a}^2\\\text{b}&\text{c}+\text{a}+\text{b}&\text{b}^2\\\text{c}&\text{a}+\text{b}+\text{c}&\text{c}^2\end{vmatrix}$
Take $(a + b + c)$ common from $C_2$​​​​​​​
$=(\text{b}+\text{c}+\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\\text{b}&1&\text{b}^2\\\text{c}&1&\text{c}^2\end{vmatrix}$
Apply$: R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1$​​​​​​​
$=(\text{b}+\text{c}+\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\\text{b}-\text{a}&0&\text{b}^2-\text{a}^2\\\text{c}-\text{a}&0&\text{c}^2-\text{a}^2\end{vmatrix}$
$=(\text{b}+\text{c}+\text{a})(\text{b}-\text{a})(\text{c}-\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\1&0&\text{b}+\text{a}\\1&0&\text{c}+\text{a}\end{vmatrix}$
$=(\text{b}+\text{c}+\text{a})(\text{b}-\text{a})(\text{c}-\text{a})(\text{b}-\text{c})$

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